Title: Part IB / Analysis and Toplogy
Tags: Topology Differentiation Metric_Spaces Convergence Compactness Continuity Contraction_Mapping Limits_And_Convergence

Q: Theorem. Any bounded sequence in \(\mathbb{R}^2\) has a convergent subsequence.
A: Let \((z_n)\) be a bounded sequence in \(\mathbb{R}^2\). Write \(z_n = (x_n, y_n)\) for each n. Now \((x_n)\) is a bounded sequence in \(\mathbb{R}\) so by Bolzano-Weierstrass has a convergent subsequence \(x_{n_j} \to x\), say.
Similarly \(y_{n_j}\) has a convergent subsequence, \(y_{n_{j_k}} \to y\), say.
But then also \(x_{n_{j_k}} \to x\), so \(z_{n_{j_k}} \to z = (x,y)\) by the fact that \[z_n \to z \iff x_n \to x, y_n \to y\]

Q: Theorem. Any Cauchy sequence in \(\mathbb{R}^2\) converges.
A: Let \(z_n = ((x_n, y_n))\) be Cauchy. For all n,m, \(|x_n - x_m|, |y_n - y_m| \leq || z_n - z_m||\).
So \((x_n), (y_n)\) Cauchy in \(\mathbb{R}\) so they converge, say \(x_n \to x, y_n \to y\), so \(z_n \to (x,y)\).

Q: Definition. \((f_n)\) converges pointwise to f, written \(f_n \to f\) pointwise if [...]
A: \(\forall x \in X,  f_n(x) \to f(x)\)
Back Extra: This doesn't feel right for some functions. Consider \(X = [0,1]\)
We can take f to be 1 at x = 0 and 0 elsewhere, now \(f_n(x) = nx, x \leq 1/n \) and 1 otherwise. Now f is not continuous, but each \(f_n\) is.

Q: Definition. Let S be a set, and f, \(f_n : S \to \mathbb{R}\) be functions. We say that \((f_n)\) converges to f uniformly on S if [...]
A: \[\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \forall x \in S, |f_n(x) - f(x)| < \epsilon\]
Back Extra: We have \(f_n \to f \text{ uniformly} \implies f_n \to f \text{ pointwise}\).

Q: Theorem. Let \(X \subseteq \mathbb{R}\), \(f_n : X \to \mathbb{R} (n \geq 1), f: X \to \mathbb{R}\). Suppose each \(f_n\) is continuous and \(f_n \to f\) uniformly. Then f is continuous.
A: Let \(x \in X\). We want to show f is continuous at X.
Let \(\epsilon > 0\), due to uniform convergence we can find N such that \[\forall y \in X, \forall n \geq N, |f_n(y) - f(y)| < \epsilon\]
In particular, \(\forall y \in X, |f_N(y) - f(y)| < \epsilon\). By continuity of \(f_N\) at x, we can find \(\delta > 0\) such that \[\forall y \in X, |y-x| < \delta \implies |f_N(y) - f_N(x)| < \epsilon\].
Now let \(y \in X\) with \(|y-x| < \delta\). We have \[|f(y) - f(x)| \leq |f(y) - f_N(y)| + |f_N(y) - f_N(x)| + |f_N(x) - f(x)| < 3\epsilon\]

Q: Theorem. Let \(f_n : [a,b] \to \mathbb{R}\) be integrable for all n. If \(f_n \to f\) uniformly on \([a,b]\), then f is integrable and \[\int_a^b f_n \to \int_a^b f\]
A: Each \(f_n\) is bounded and \(f_n \to f\) uniformly so for all n sufficiently large, \(f_n - f\) is bounded. Hence \(f = f_n - (f_n - f)\) is bounded.
Let \(\epsilon > 0\). As \(f_n \to f\) uniformly we can find N such that \[\forall x \in [a,b], |f_N(x) - f(x)| < \epsilon\]
By Riemann's criterion, as \(f_N\) is integrable we can find a dissection \(\mathcal{D}\) such that \(S(f_n, \mathcal{D}) - s(f_N, \mathcal{D}) < \epsilon\).
Write \(\mathcal{D} = \{x_0 < x_1 < \dots < x_k\}, x_0 = a, x_k = b\).
Using definition of upper sum, and observing that for any \(I \subset [a,b], \sup_I f \leq \sup_I f_N + \epsilon\). Hence \(S(f, \mathcal{D}) \leq S(f_n, \mathcal{D}) + (b-a)\epsilon\). Now, \(S(f, \mathcal{D}) - s(f, \mathcal{D}) \leq S(f_N, \mathcal{D}) - s(f_N, \mathcal{D}) + 2(b-a)\epsilon < (2(b-a) + 1)\epsilon\). So by Riemann's criterion, f is integrable.
Taking n sufficiently large that \(f_n - f\) is bounded, we have \[\left | \int_a^b f_n - \int_a^b f \right | \leq \int_a^b |f_n - f| \leq (b-a)\sup_{[a,b]} |f_n - f| \to 0\] as \(f_n \to f\) uniformly.

Q: Theorem. Let \(f_n : (u,v) \to \mathbb{R} (n \geq 1)\) be continuously differentiable. Let \(f : (u,v) \to \mathbb{R}\) with \(f_n \to f\) pointwise. Suppose \(g : (u,v) \to \mathbb{R}\) with \(f_n' \to g\) uniformly, then f is differentiable and \(f' = g\).
A: Fix some \(a \in (u,v)\). Let \(x \in (u,v)\) be arbitrary. By FTC for each n we have \(f_n(x) = f_n(a) + \int_a^x f_n\).
We have \(f_n \to f\) pointwise and \(f_n' \to g\) uniformly so we have \(\int_a^x f_n' \to \int_a^x g\). Hence \(f(x) = f(a) + \int_a^x g\).
We also have \(g\) is continuous so by FTC again, f is differentiable at x and \(f'(x) = g(x)\).

Q: Definition. Uniformly Cauchy. Let \(X \subset \mathbb{R}, f_n : X \to \mathbb{R} (n \geq 1)\). We say that \((f_n)\) is uniformly Cauchy if [...]
A: \[\forall \epsilon > 0, \exists N \text{ such that } \forall x \in X, \forall m,n \geq N, |f_m(x) - f_n(x)| < \epsilon\]
Back Extra: Uniform convergence implies uniform Cauchy.

Q: Theorem. General Principle of Uniform Convergence. Let \(X \subset \mathbb{R}\) and let \(f_n : X \to \mathbb{R} (n \geq 1)\). Suppose \((f_n)\) is uniformly Cauchy. Then it is uniformly convergent.
A: i)We can show \(f_n \to f\) point wise. Let \(x \in X\) and \(\epsilon > 0\). As \((f_n)\) is uniformly Cauchy, there is some N s.t. \(\forall y \in X, \forall m,n \geq N, |f_m(y) - f_n(y)| < \epsilon\). That is \((f_n(x))\) is Cauchy in \(\mathbb{R}\). So by general principle of convergence, \(f_n(x) \to f(x)\).
ii) Next we show \(f_n \to f\) uniformly. Let \(\epsilon > 0\), as \((f_n)\) is uniformly Cauchy, for some N \(\forall x \in X, \forall m,n \geq N, |f_m(x) - f_n(x)| < \epsilon\). Let x be arbitrary, fix m and let n go to infty. By point wise convergence we have \(\forall m \geq N |f_m(x) - f(x)| \leq \epsilon\),  but x was arbitrary so \[\forall x \in X \forall m \geq N |f_m(x) - f(x)| \geq \epsilon\].
Hence \(f_n \to f \) uniformly.

Q: Definition. Let \(X \subset \mathbb{R}, f_n : X \to \mathbb{R} (n \geq 1)\). We say \((f_n)\) is pointwise bounded if [...]
We say \((f_n)\) is uniformly bounded if [...]
A: \[\forall x \in X,  \exists M \text{ s.t. } \forall n |f_n(x)| \leq M\]; \[\exists M \text{ s.t. } \forall x \in X, \forall n |f_n(x)| \leq M\]

Q: Lemma. Let \(\sum_n a_n x^n\) be a real power series with radius of convergence R. Let \(0 < r < R\). Then \(\sum_n a_n x^n\) converges uniformly on \((-r,r)\)
A: Define \( f : (-r,r) \to \mathbb{R} \) by \(f(x) = \sum_{n=0}^{\infty} a_n x^n\) and, for \(m \geq 1\), \(f_m(x) = \sum_{n=0}^m a_n x^n\).
For any \(m,x,\) we have \[|f_m(x) - f(x)| = |\sum_{n = m+1}^{\infty}a_n x^n| \leq \sum_{n = {m+1}^{\infty}} |a_n||x|^n \leq \sum_{n = m+1}^{\infty} |a_n| r^n\] Hence for any \(m \geq 1\), \( f_m - f\) is bounded and \[\sup_{(-r,r)} |f_m - f| \leq \sum_{n = m+1}^{\infty} |a_n|r^n \to 0\] as \(m \to \infty\) by absolute convergence of \(\sum a_n r^n\), hence \(f_m \to f\) uniformly.

Q: Lemma. Let \(\sum_n a_n x^n\) be a real power series with r.o.c R. Then the power series \(\sum_n na_n x^{n-1}\) also has radius of convergence R.
A: Exercise bruh look it up.

Q: Theorem. Let \(\sum_n a_n x^n\) be a real power series with r.o.c R. Define \(f : (-R, R) \to \mathbb{R}\) by \(f(x) = \sum_{n=0}^{\infty} a_n x^n\). Then
i) f is continuous;
ii) for any \(x \in (-R,R)\), f is integrable over \([0,x]\) with \[\int_0^x f = \sum_{n=0}^{\infty} \frac{1}{n+1} a_n x^{n=1}\]
iii)f is a differentiable function and for all \(x \in (-R,R)\), \(f'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}\)
A: For each \(m \geq 1\), define \(f_m : (-R,R) \to \mathbb{R}\) by \[f_m(x) = \sum_{n=0}^m a_n x^n\]
Let \(x \in (-R,R)\) be arbitrary. Choose r with \(|x| < r_m < R\). Then by lemma 8, \(f_m \to f\) uniformly on \((-r,r)\).
i) The function \(f|_{(-r,r)}\) is a continuous function as \(f_m \to f\) uniformly on \((-r,r)\) and each \(f_m\) is a polynomial so continuous. In particular \(f|_{(-r,r)}\) is continuous at x.
But \(x \in (-R,R)\) was arbitrary so f is a continuous function.
ii) We have \([0,x] \subset (-r,r)\). For each m, we know \(f_m\) is integrable over \([0,x]\) and \(\int_0^x f_m = \sum_{n=0}^m \frac{1}{n+1} a_n x^{n+1}\)/ We know f is integrable over \([0,x]\) and \[\int_0^x f = \lim_{m \to \infty} \int_0^x f_m = \sum_{n=0}^{\infty} \frac{1}{n+1}a_n x^n\]
iii) For each m, \(f_m\) is a polynomial so \(f_m\) is differentiable on \((-r,r)\) and for each \(y \in (-r,r)\), \(f_m'(y) = \sum_{n=1}^m na_n y^{n-1}\). By our lemma, \(\sum na_n y^{n-1}\) has r.o.c at least R, so by another lemma, the series \(\sum_{n \geq 1} na_ny^{n-1}\) converges uniformly on \((-r,r)\). Hence by our theorem, \(f|_{(-r,r)}\) is a differentiable function with \(\forall y \in (-r,r), f'(y) = \sum_{n=1}^{\infty} na_n y^{n-1}\). Hence f is differentiable at x with \(f'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}\)

Q: Definition. We say f is uniformly continuous if [...].
A: \[\forall \epsilon > 0 \; \exists \delta > 0 \; \forall x,y \in X |x-y| < \delta \implies |f(x) - f(y)| < \epsilon\]

Q: Theorem. Let \(f: [a,b] \to \mathbb{R}\) be continuous. Then f is uniformly continuous.
A: Suppose f is not uniformly continuous. Then we can find \(\epsilon > 0\) s.t. \[ \forall \delta > 0 \; \exists x,y \in [a,b] \; |x-y| < \delta \text{ but } |f(x) - f(y)| \geq \epsilon \]
Taking \(\delta = 1/n\) in the above we obtain sequences \((x_n), (y_n)\) in \([a,b]\) such that for each n, \(|x_n - y_n| < 1/n\) but \(|f(x_n) - f(y_n)| \geq \epsilon\).
The sequence \((x_n)\) is bounded so by BW it has a convergent subsequence, say \(x_{n_j} \to x\). Moreover, \(\forall j \; a \leq x_{n_j} \leq b\) so \(a \leq x \leq b\). As f is continuous, here \(f(x_{n_j})\to f(x)\) as \(j \to \infty\).
Thus \(f(x_{n_j}) - f(y_{n_j}) \to 0\) as \(j \to \infty\). But \(\forall j |f(x_{n_j}) - f(y_{n_j})| \geq \epsilon\). Contradiction.
Hence f is uniformly continuous.

Q: Corollary: Let \(f:[a,b] \to \mathbb{R}\) be continuous. Then f is bounded.
A: As f is continuous, and the interval is closed and bounded, we know f is uniformly continuous. Hence we can find some \(\delta > 0\) such that \(\forall x,y \in [a,b], |x-y| < \delta \implies |f(x) - f(y)| < 1\). Let \(M = \lceil \frac{b-a}{\delta} \rceil +1\). Suppose \(x \in [a,b]\). Then we can find \(a = x_0 \leq x_1 \leq x_2 \leq \dots \leq x_M = x\) with \(|x_i - x_{i-1}| < \delta\) for every i. Then \(f(x) = |f(a) + \sum_{i=1}^m f(x_i) - f(x_{i-1})| \leq |f(a)| + M\)

Q: Corollary. Let \(f:[a,b] \to \mathbb{R}\) be continuous. Then f is integrable.
A: f is uniformly continuous as it is continuous and defined on a closed bounded interval. Let \(\epsilon > 0\). Then we can find a \(\delta > 0\) with \(\forall x,y \in [a,b], |x-y| < \delta \implies |f(x) - f(y)| < \epsilon\). Let \(\mathcal{D} = \{x_0 < x_1 < \dots < x_n\}\) be a dissection with \(x_i - x_{i-1} < \delta\) for each i. For each i, if \(x,y \in [x_{i-1}, x_i]\) then \(|x-y| < \delta\) and so \(|f(x) - f(y)| < \epsilon\).
Hence for each i, \(\sup_{[x_{i-1}, x_i]} f - \inf_{[x_{i-1}, x_i]} f \leq \epsilon\).
Hence \(S(f, \mathcal{D}) - s(f, \mathcal{D}) = \sum_{i=1}^n (x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]} f - \inf_{[x_{i-1}, x_i]} f) \leq \sum_{i=1}^n (x_i - x_{i-1}) \epsilon = (b-a)\epsilon\).
Hence by Riemann's Criterion, f is integrable over \([a,b]\).

Q: Definition. A metric space is a set X endowed with a metric, i.e. a function \(d : X^2 \to \mathbb{R}\) with the following properties: [...]
A: i) \(\forall x,y \in X \; \; d(x,y) \geq 0\)  with equality iff \(x = y\);
ii) \(\forall x,y \in X \; \; d(x,y) = d(y,x)\);
iii) \(\forall x,y,z \in X \; \; d(x,z) \leq d(x,y) + d(y,z)\)

Q: Definition. Let \(X = B(A) = \{f: A \to \mathbb{R} | \text{f bounded}\}\) with the uniform metric [...]
A: \[d(f,g) = \sup_A |f-g|\]
Back Extra: Easy to show metric properties hold.

Q: Definition. Let \(X,d\) be a metric space and \(Y \subset X\). Then \(d |_{y^2}\) is clearly a metric on Y. We say Y with this restricted metric is [...]
A: a subspace of \((X,d)\).
Back Extra: For example any subset of \(\mathbb{R}\) with the usual metric.

Q: Definition. Let \((X,d)\) be metric space, let \((x_n)\) be a sequence in X and let \(x \in X\). We say \((x_n)\) converges to x and write \(x_n \to x\) if [...].
A: \(d(x_n, x) \to 0 \)

Q: Proposition. Limits are unique. That is, let \((X, d)\) be a metric space, \((x_n)\) a sequence in X and \(x,y \in X\) with \(x_n \to x\) and \(x_n \to y\). Then \(x = y\).
A: We have for each n, by triangle inequality, \[d(x,y) \leq d(x, x_n) + d(x_n, y) = d(x_n , x) + d(x_n, y) \to 0 + 0 = 0\] by results in \(\mathbb{R}\).

Definition. Let \((X,d)\) and \((Y,e)\) be metric spaces and let \(f: X \to Y\) be a function.
i) Let \(a \in X\) and \(b \in Y\). We say \(f(x) \to b\) as \(x \to a\) if \[\forall \epsilon > \; \exists \delta > 0 \; \forall x \in X \; 0 < d(x,a) < \delta \implies e(f(x),b) < \epsilon\]
ii) Let \(a \in X\). We say f is continuous at a if \(f(x) \to f(a)\) as \(x \to a\); that is \[\forall \epsilon > 0 \; \exists \delta > 0 \; \forall x \in X \; d(x,a) < \delta \implies e(f(x), f(a)) < \epsilon\]
iii) We say f is a continuous function if it is continuous at every point in X.
iv) Let \(W \subset X\). We say f is continuous on W if \(f|_W : W \to Y\) is a continuous function, with W thought of as a subspace of X; that is \[\forall a \in W \; \forall \epsilon > 0 \; \exists \delta > 0 \; \forall x \in W \; \; d(x,a) < \delta \implies e(f(x), f(a)) < \epsilon\]
v) We say f is uniformly continuous if \[\forall \epsilon > 0 \; \exists \delta > 0 \; \forall x,y \in X \; d(x,y) < \delta \implies e(f(x), f(y)) < \epsilon\]

Q: Proportion. Let \((X,d)\) and \((Y,e)\) be metric spaces. Let \(f : X \to Y\). Let \(a \in X\). Then f is continuous at a iff whenever \((x_n)\) is a sequence in X with \(x_n \to a\) then \(f(x_n) \to f(a)\)
A: (->) Assume f is continuous at a. Let \((x_n)\) be a sequence in X with \(x_n \to a\). Let \(\epsilon > 0\). As f is continuous at a, we can find a \(\delta > 0\) s.t. \(\forall x \in X \; \; d(x,a) < \delta \implies e(f(x), f(a)) < \epsilon\). And as \(x_n to a\), we can find an N s.t. \(\forall n \geq N \; \; d(x_n , a) < \delta\). Now let \(n \geq N\). Then \(d(x_n , a) < \delta\) and so \(e(f(x_n), f(a)) < \epsilon\). Hence \(f(x_n) \to f(a)\).
(<-) Suppose f is not continuous at a. Then there is some \(\epsilon > 0\) s.t. \(\forall \delta > 0 \; \exists x \in X\) s.t. \(d(x, a) < \delta\) but \(e(f(x), f(a)) \geq \epsilon\). Taking \(\delta = \frac{1}{n}\) in this statement, for each n we have a point \(x_n \in X\) with \(d(x_n, a) < \frac{1}{n}\) but \(e(f(x_n), f(a)) \geq \epsilon\). Hence \(x_n \to a\) but \(f(x_n) \not \to f(a)\)

Q: Proposition. Let \((W,c), (X,d), (Y,e)\) be metric spaces. Let \(f : W \to X\) and \(g : X \to Y\) be continuous at \(a \in W\) and \(f(a) \in X\) respectively. Then the composition \(g \circ f : W \to Y\) is continuous at a.
A: Let \((x_n)\) be a sequence in W with \(x_n \to a\). As f is continuous at a we have \(f(x_n) \to f(a)\), and as g is continuous at \(f(a)\) we have \(g(f(x_n)) \to g(f(a))\). Therefore \(g \circ f \) is continuous at a.

Q: Definition. Let \((X,d)\) be a metric space and let \((x_n)\) be a sequence in X. We say \((x_n)\) is a Cauchy sequence if [...]. We say that X is complete if [...]
A: \(\forall \epsilon > 0 \; \exists N \; \forall m,n \geq N \; d(x_m , x_n) < \epsilon\); every Cauchy sequence in X is convergent.
Back Extra: Easy to show convergent implies Cauchy.

Q: Definition. Let \((X,d)\) be a metric space and \(Y \subset X\). We say Y is closed if [...]
A: whenever \((x_n)\) is a sequence in Y with \(x_n \to x\) in X then in fact \(x \in Y\).

Q: Proposition. Let X be a complete metric space and \(Y \subset X\). Then \[Y \text{ is complete} \iff Y \text{ is a closed subset of X}\]
A: (<-) Suppose Y is closed. Let \((x_n)\) be a Cauchy sequence in Y. As X complete, \((x_n)\) converges in X, say \(x_n \to x \in X\). As Y is closed, \(x \in Y\).
(->)Suppose Y complete. Let \((x_n)\) be a sequence in Y with \(x_n \to x\) in X. Then \((x_n)\) is convergent in X so \((x_n)\)is Cauchy so, as Y is complete, \((x_n)\) is also convergent in Y, say \(x_n \to y \in Y\). But then \(x_n \to y\) in X so by uniqueness of limits in X we have \(x = y\). Hence \(x \in Y\).

Q: Definition. Let \((X,d)\) be a metric space and let \(f : X \to X\). We say f is a contraction if [...]
A: there is some constant \(\lambda \in [0,1)\) s.t \[\forall x,y \in X \; d(f(x), f(y)) \leq \lambda d(x,y)\]
Back Extra: Obviously continuous and indeed uniformly continuous.

Q: Theorem. Contraction Mapping Theorem. Let \((X,d)\) be a non-empty complete metric space. Let \(f : X \to X\) be a contraction. Then f has a unique fixed point.
A: As \(X \neq \emptyset\) , can pick a point \(x_0 \in X\). As f is a contraction, then there is some \(\lambda \in [0,1)\) such that \[\forall x,y \in X \; d(f(x), f(y)) \leq \lambda d(x,y)\].
Recursively define \(x_{n+1} = f(x_n)\) for \(n \geq 0\). Let \(\Delta = d(x_0 , x_1)\). Then, by induction, for all \(n \geq 0\) we have \(d(x_n , x_{n+1}) \leq \lambda^n \Delta\). Hence if \(N \leq m < n\) we have \[d(x_m , x_n) \leq \sum_{i=m}^{n-1} d(x_i , x_{i+1}) \leq \sum_{i=m}^{n-1} \lambda^i \Delta \leq \sum_{i=N}^{\infty} \lambda^i \Delta = \frac{\lambda^N \Delta}{1- \lambda} \to 0 \text{ as } N \to \infty \]. Hence \((x_n)\) is Cauchy, as X is complete the sequence converges, say \(x_n \to x \in X\). Then also \(x_{n+1} \to x\). But f is continuous so \(x_{n+1} = f(x_n) \to f(x)\). SO by uniqueness of limits, \(f(x) = x\).
Suppose also \(y \in X\) with \(f(y) = y\). Then \[d(x,y) = d(f(x), f(y)) \leq \lambda d(x,y)\] where \(\lambda < 1\) so \(d(x,y) = 0\) and so \(y = x\).

Q: Lemma. Let \(U \subset \mathbb{R}^n\) is closed and bounded and \(V \subset \mathbb{R}^m\)is closed. Then any continuous function \(U \to V\) is bounded and moreover, the metric space \[C(U,V) = \{f : U \to V | f \text{ continuous}\}\] with uniform metric is complete.
A: Will see later.

Q: Theorem. Picard-Lindelöf. Let \(a,b \in \mathbb{R}\) with \(a < b\) and let \(t_0 \in (a,b)\). Let \(y_0 \in \mathbb{R}^n\) and let \(\delta > 0\). Let \(\phi : [a,b] \times B_{\delta}(y_0) \to \mathbb{R}^n\) be continuous such that there exists some K>0 such that \(\forall t \in [a,b], \forall x,y \in B_{\delta}(y_0), || \phi(t,x) - \phi(t,y)|| \leq K|| x - y||\).
Then there exists some \(\epsilon > 0\) s.t. \([t_0 - \epsilon, t_0 + \epsilon] \subset [a,b]\) and the IVP \[f'(t) = \phi(t, f(t)\] with \(f(t_0) = y\) has a unique solution on \([t_0 - \epsilon, t_0 + \epsilon]\)
A: We know \(\phi\) is bounded from a lemma. So we can pick M s.t. \(\forall t \in [a,b], \forall x \in B_{\delta}(y_0) \; || \phi(t,x)|| \leq M\). Let \(\epsilon > 0\) with \([t_0 - \epsilon, t_0 + \epsilon] \subset [a,b]\) to be specified later. Let \(X = C([t_0 - \epsilon, t_0 + \epsilon] , B_{\delta}(y_0))\). So again by the same Lemma, X is a complete metric space with the uniform metric. For \(f \in X\) and \(t \in [t_0 - \epsilon, t_0 + \epsilon]\), define \[T f(t) = y_0 + \int_{t_0}^t \phi(s,f(s)) ds\]
Now for any \(f \in X\) and any \(t \in [t_0 - \epsilon, t_0 + \epsilon]\), \[|| Tf(t) - y_0 || \leq \int_{t_0}^t || \phi (s, f(s)) || ds \leq M || t - t_0 || \leq M \epsilon \leq \delta\] as long as \(\epsilon \leq \delta / M\).
Hence as long as we choose a good enough \(\epsilon\) we have T maps \(X \to X\). Next for any \(f,g \in X\) and \(t \in [t_0 - \epsilon, t_0 + \epsilon]\) we have \[|| Tf(t) - T g(y) || \leq \int_{t_0}^t || \phi (s,f(s)) - \phi(s,g(s)) || ds \leq \int_{t_0}^t K || f(s) - g(s) || ds \leq |t - t_0| K d(f,g) \leq K \epsilon d(f,g) \leq 1/2 d(f,g)\] as long as we pick \(\epsilon \leq 1/2K\). So choose \(\epsilon\) to be the minimum of 1/2K and \(\delta / M\). Then \(T : X \to X\) and for any \(f, g \in X\) we have \(d(Tf, Tg) \leq 1/2 d(f,g)\) and so T is a contraction of X.
So by the Contraction Mapping Theorem, T has a unique fixed point f. And by FTC, this is precisely the same as a solution of the IVP.

Q: Definition. Let \((X,d)\) be a metric space.  We say that X is sequentially compact if [...]
A: every sequence in X has a convergent subsequence.

Q: Theorem. Let \(X \subset \mathbb{R}^n\). Then \[X \text{ sequentially compact} \iff X \text{ closed and bounded}\]
A: (<-) Suppose X closed and bounded. Let \((x_n)\) be a sequence in X. Then by BW for \(\mathbb{R}^n\), \((x_n)\) has a convergent subsequence \(x_{n_j} \to x \in \mathbb{R}^n \). But X is closed so \(x \in X\). So X is sequentially compact.
(->) First suppose X not closed. Then there is a sequence \((x_n)\) in X with \(x_n \to x \in \mathbb{R}^n\) but \(x \not \in X\). Suppose \((x_{n_j})\) is a subsequence with \(x_{n_j} \to y \in X\). Then \(x_{n_j} \to y\) in \(\mathbb{R}^n\) so y = x. But \(x \not \in X\). Contradiction.
Suppose instead X is not bounded. Then we have a sequence \((x_n)\) in X with \(||x_n || \to \infty\). Suppose \((x_{n_j})\) is a subsequence with \(x_{n_j} \to y \in X\). Then \(||x_{n_j}|| \to ||y||\) but \(||x_{n_j} \to \infty\). Contradiction

Q: Definition. A metric space (X,d) is bounded if [...]
A: \(\exists M \; \forall x,y \in X \; d(x,y) \leq M\)
Back Extra: Equivalently, X is bounded iff X is empty or \(\exists x \in X \; \exists K \; \forall y \in X \; d(x,y) \leq K\).

Q: Definition. Let \((X,d)\) be a metric space. We say X is totally bounded if [...]
A: for every \(\epsilon > 0\) there is a finite \(A \subset X\) s.t. \(\forall x \in X \; \exists a \in A \; d(x,a) < \epsilon\)
Back Extra: X totally bounded implies X bounded. But converse is not true.

Q: Theorem. Let \((X,d)\) be a metric space. Then X is sequentially compact iff X is complete and totally bounded.
A: (<-) Suppose X is complete and totally bounded, and let \((x_n)_{n \geq 1}\) be a sequence in X. By total boundedness, we can define a finite \(A \subset X\) satisfying \(\forall x \in X \; \exists a \in A \; d(x,a) < \frac{1}{2}\). In particular, there is some \(a \in A\) s.t. \(d(x_n , a)< \frac{1}{2}\) for infinitely many n. In other words, \((x_n)_{n \geq 1}\) has a subsequence \((x_{n_{1,j}})_{j \geq 1}\) s.t. \(\forall j d(x_{n_{1,j}}) < \frac{1}{2}\) and so \(\forall j,k \; d(x_{n_{1,j}} , x_{n_{1,k}}) < 1\). Similarly, \((x_{n_{1,j}})_{j \geq 1}\) has a subsequence \((x_{n_{2,j}})_{j \geq 1}\) satisfying \[ \forall j,k \; d(x_{n_{2,j}} , x_{n_{2,k}}) < \frac{1}{2} \]
Repeating this we get a subsequence \((x_{n_{i,j}})_{j \geq 1}\) of \((x_n)_{n \geq 1}\) satisfying
i) for every \(i \geq 1\), \( (x_{n_{i+1, j}})_{j \geq 1}\) is a subsequence of \((x_{n_{i,j}})_{j \geq 1}\), and
ii) \(\forall i,j,k \; d(x_{n_{i,j}} , x_{n_{i,k}}) < \frac{1}{i}\)
Observe that \(n_{1,1} < n_{2,2} < \dots\) so \(( x_{n_{i,i}})_{i \geq 1}\) is a subsequence of \((x_n)_{n \geq 1}\). Moreover, if \(i \leq j\) then \(d(x_{n_{i,i}} , x_{n_{j,j}}) \geq \frac{1}{i}\). Thus \((x_{n_{i,i}})_{i \geq 1}\) is a Cauchy sequence and hence, by completeness, converges. So X is sequentially compact.
(->) Suppose first that X is not complete. Then X contains a Cauchy sequence that does not converge. Suppose \((x_n)\) is the sequence and has a convergent subsequence, say \(x_{n_j} \to x\). Then, as \((x_n)\) Cauchy, \(x_n \to x\) which is a contradiction so X is not sequentially compact.
Suppose now instead that X is not totally bounded. Then we can find \(\epsilon > 0\) s.t. there is no finite set \(A \subset X\) with \(\forall x \in X \; \exists a \in A \; d(x,a) < \epsilon\). In particular, \(X \neq \emptyset \) so pick \(x_1 \in X\). Recursively, given \(x_1 , \dots x_n \in X\), we can pick \(x_{n+1} \in X\) with \(d(x_i , x_{n+1}) \geq \epsilon\) for all \(i = 1,2, \dots , n\). Hence we obtain a sequence \((x_n)\) in X with \(\forall i,j \; i \neq j \implies d(x_i , x_j) \geq \epsilon\). In particular, \((x_n)\) can have no convergent subsequence. So X is not sequentially compact.

Q: Theorem. Let \((X,d), (Y,e)\) be metric spaces with X sequentially compact and let \(f:X \to Y\) be continuous. Then f is uniformly continuous and \(f(X)\) is bounded. Moreover, if \(Y\) is \(\mathbb{R}\) with the usual metric then there are \(s,t \in X\) with \(f(s) = \inf_{x \in X} f(x)\) and \(f(t) = \sup_{x \in X} f(x)\)
A: Suppose f is not uniformly continuous. Then there is some \(\epsilon > 0\) s.t. \(\forall \delta > 0 \; \exists x,y \in X \; d(x,y) < \delta\) but \(e(f(x), f(y)) \geq \epsilon\). In particular, taking \(\delta = 1/n\) for \(n = 1,2,\dots\) we get sequences \((x_n)\) and \((y_n)\) in X s.t. for all n we have \(d(x_n , y_n) < 1/n \) but \(d(f(x_n) , f(y_n)) \geq \epsilon\). By sequential compactness, \((x_n)\) has a convergent subsequence, say \(x_{n_j} \to x\). But \(d(x_{n_j}, y_{n_j}) \to 0 \) so also \(y_{n_j} \to x\). And f is continous so \(f(x_{n_j}) \to f(x)\) and \(f(y_{n_j}) \to f(x)\) giving \(e(f(x_{n_j}), f(y_{n_j})) \to 0\). But \(\forall j \; e(f(x_{n_j}) , f(y_{n_j})) \geq \epsilon\). This is a contradiction.
Suppose \(f(X)\) is not bounded. Then we can find sequences \((x_n), (y_n)\) in X such that \(e(f(x_n) , f(y_n)) \to \infty\) as \(n \to \infty\). By sequential compactness, \((x_n)\) has a convergent subsequence, say \(x_{n_j} \to x\), and then \((y_{n_j})\) has a convergent subsequence, say \(y_{n_{j_k}} \to y\). Also \(x_{n_{j_k}} \to x\). As f is continuous, \(f(x_{n_{j_k}}) \to f(y)\) and so \(e(f(x_{n_{j_k}}), f(y_{n_{j_k}})) \to e(f(x), f(y))\). But \(e(f(x_{n_{j_k}}) , f(y_{n_{j_k}})) \to \infty\). Contradiction.
Finally, let \(Y = \mathbb{R}\) and e be the usual metric on \(\mathbb{R}\). We know f is bounded so let \(\tau = \inf_{x \in X} f(x)\). Suppose this infimum is not attained. We can then define \(g : X \to \mathbb{R}\) by \(g(x) = \frac{1}{f(x) - \tau}\). Then g is a continuous \(f_n\) on X but \(g(X)\) is not bounded. Contradiction.
Similarly, the supremum is attained

Q: Corollary. Let \((X,d)\) be a metric space, let \(K \subset X\) be sequentially compact and let \(x \in X \setminus K\). Then there is some \(\epsilon > 0\) s.t. \(\forall y \in K \; d(x,y) \geq \epsilon\) and \(\exists z \in K \; d(x,z) = \epsilon\).
A: We can define \(y \mapsto d(x,y)\) defining a continuous function on \(K \to \mathbb{R}\) which is always strictly positive, and therefore it is bounded and attains its infimum.

Q: Definition. Let \((X,d), (Y,e)\) be metric spaces and let \(f: X \to Y\). We say f is an isometry if [...]
A: \[\forall x,y \in X \; e(f(x), f(y)) = d(x,y)\]
Back Extra: Clearly an isometry is a continuous injection. But doesn't have to be surjective( e.g. X proper subspace of Y and \(f(x) = x\) for all x ) even if X = Y (e.g. \(X = [0 , \infty)\) and \(f(x) = x+1\) for all x)

Q: Theorem. Let \((X,d)\) be a sequentially compact metric space and \(f: X \to X\) an isometry. Then f is bijective.
A: Suppose not. Then f is not surjective so \(\exists x \in X\) s.t. \(x \not \in f(X)\). Now, f is continuous and X is sequentially compact so can check that \(f(X)\) is sequentially compact (we will see this later). So there is some \( \epsilon > 0 \) such that \(\forall y \in X \; d(x, f(y)) \geq \epsilon\).
Now let \(x_0 = x\) and recursively define \(x_{n+1} = f(x_n)\). Then \[\forall n \geq 1 d(x_0 , x_n) \geq \epsilon\].
Moreover, as f is an isometry, if \(i < j\) then \(d(x_i , x_j) = d(x_0 , x_{j - i}) \geq \epsilon\). So \(\forall m,n \; m \neq n \implies d(x_m, x_n) \geq \epsilon\). So \((x_n)\) has no convergent subsequence. Contradiction.

Q: Definition. Let \((X,d), (Y,e)\) be metric spaces. A homeomorphism from X to Y is [...] If such a homeomorphism exists, we say X and Y are homeomorphic. If identity map from \((X,d)\) to \((X, d')\) is a homeomorphism, we say [...]
A: a continuous bijection \(f: X \to Y\) with continuous inverse.; d,d' are equivalent.

Q: Proposition. Let \((V,b),(W,c), (X,d), (Y,e)\) be metric spaces. Let \(g:X \to V\) and \(h : Y \to W\) be homeomorphisms. Then
i) In X, \(x_n \to x\) iff in V \(g(x_n) \to g(x)\);
ii) a function \(f : X \to Y\) is continuous at \(a \in X\) iff \(h \circ f \circ g^{-1} : V \to W\) is continuous at g(a).
A: i) \(x_n \to x \implies g(x_n) \to g(x)\) as g continuous.
\(g(x_n) \to g(x) \implies x_n = g^{-1}(g(x_n)) \to g^{-1}(g(x)) = x\) as \(g^{-1}\) continuous.
ii) f continuous at a implies \(h \circ f \circ g^{-1}\) is continuous at g(a) as it is a composition of continuous functions.
If \(h \circ f \circ g^{-1}\) is continuous at g(a), then \(f = h^{-1} \circ (h \circ f \circ g^{-1}) \circ g\) is continuous at a similarly.

Q: Definition. Let \((X,d)\) be a metric space, \(a \in X\) and \(\delta > 0\). Then the open ball of radius \(\delta\) about \(a\) is the set [...]
A: \[B_{\delta}(a) = \{x \in X | d(x,a) < \delta\}\]
Back Extra: Suppose X,Y are metric spaces and \(f: X \to Y, a \in X\). Then f continuous at a iff \(\forall \epsilon > 0 \; \exists \delta > 0 \; d(x,a) < \delta \implies e(f(x), f(a)) < \epsilon\). This is the same as \[\forall \epsilon > 0 \; \exists \delta > 0 \; x \in B_{\delta}(a) \implies f(x) \in B_{\epsilon}(f(a)) \iff \forall \epsilon > 0 \; \exists \delta > 0 \; f(B_{\delta}(a)) \subset B_{\epsilon}(f(a)) \iff \forall \epsilon > 0 \; \exists \delta > 0 \; B_{\delta}(a) \subset f^{-1}(B_{\epsilon}(f(a)))\]

Q: Definition. Let X be a metric space. We say \(G \subset X\) is open if [...] We say \(\mathcal{N} \subset X\) is a neighbourhood of \(a \in X\) if [...]
A: for all \(a \in G\) there is some \(\delta > 0\) s.t. \(B_{\delta}(a) \subset G\).; there is some open \(G \subset X\) s.t. \(a \in G \subset \mathcal{N}\).

Q: Proposition. Let \((X,d)\) and \((Y,e)\) be metric spaces and let \(f: X \to Y\).
i) f is continous at a point \(a \in X\) iff whenever N is a neighbourhood of \(f(a)\) in Y then \(f^{-1}(N)\) is a neighbourhood of a in X
ii) f is a continuous function iff whenever \(G \subset Y\)is open then \(f^{-1}(a) \subset X\) is open.
A: i) (->) Let \(N \subset Y\) be a neighbourhood of f(a). Then we have some \(\epsilon > 0\) s.t. \(B_{\epsilon}(f(a)) \subset N\). But f is continuous at a so we have some \(\delta > 0\) s.t. \(B_{\delta}(a) \subset f^{-1}(B_{\epsilon}(f(a)))\) and so \(B_{\delta}(a) \subset f^{-1}(N)\). So \(f^{-1}(N)\) is a neighbourhood of a.
(<-) Let \(\epsilon > 0\). Then \(B_{\epsilon}(f(a))\) is a neighbourhood of f(a). So \(f^{-1}(B_{\epsilon}(f(a)))\) is a neighbourhood of a. So we have some \(\delta > 0\) s.y. \(B_{\delta}(a) \subset f^{-1}(B_{\epsilon}(f(a)))\). So f is continuous at a.
ii) (->) Let \(G \subset Y\) be open. Let \(a \in f^{-1}(G)\). Then \(f(a) \in G\) and G open so G is a neighbourhood of \(f(a)\). And f is continuous at a so by (i), \(f^{-1}(a)\) is a neighbourhood of a. Then a was arbitrary, so \(f^{-1}(a)\) is a neighbourhood of a. Then a was arbitrary, so \(f^{-1}(a)\) is a neighbourhood of each of its points so \(f^{-1}(a) \subset X\) open.
(<-) Let \(a \in X\) be arbitrary. Let \(N \subset Y\) be a neighbourhood of \(f(a)\). Then there is an open set \(G \subset Y\) s.t. \(f(a) \in G \subset N\). By assumption, \(f^{-1}(G) \subset X\) is open. SO we have \(a \in f^{-1}(G) \subset f^{-1}(N)\) with \(f^{-1}(G)\) open so \(f^{-1}(N)\) is a neighbourhood of a. So by (i), f is continuous at a. But a was arbitrary so f is a continuous function.

Q: Definition. Let \((X,d)\) be a metric space. An open cover of X is [...] A subcover of C is [...] We say X is compact if [...]
A: a collection C of open sets in X such that \(\bigcup_{G \in C} G = X\).; a \(B \subset C\) which is itself an open cover of X.; every open cover of X has a finite subcover.

Q: Theorem. Let \((X,d)\) be a metric space. Then TFAE:
i) X is compact
ii) X is sequentially compact
iii)X is complete and totally bounded
and if X is a subspace of \(\mathbb{R}\) with the usual metric
iv) \(X \subset \mathbb{R}\) is closed and bounded.
A: We already know \((ii) \iff (iii)\) and \((ii) \iff (iv)\) . So it suffices to prove \((i) \iff (ii)\)
(i -> ii)Suppose X is compact but not sequentially compact. So there is a sequence \((x_n)\) in X with no convergent subsequence. Then for each \(a \in X\) then there is an open set \(G_a\) with \(a \in G_a\) s.t. \(x_n \in G_a\) for only finitely many values of n.
(Otherwise pick \(n_1\) s.t. \(x_{n_1} \in B_1(a)\), then \(n_2 > n_1\) with \(x_{n_2} \in B_{1/2}(a)\), again and again giving \(x_{n_j} \to a\)\)).
Let \(\mathcal{C} = \{G_a | a \in X\}\). Then \(\mathcal{C}\) is an open cover of X so has a finite subcover, say \(\{ G_{a_1} , \dots , G_{a_n}\}\). But then \(X = \cup_{i=1}^n G_{a_i}\) contains \(x_n\) for only finitely many values of n. Contradiction.
(ii -> i) Suppose X is sequentially compact. Let \(\mathcal{C}\) be an open cover of X. We first show there is some \(\delta > 0\) s.t. \(\forall a \in X \; \exists G \in \mathcal{C} \; B_{\delta}(a) \subset G\). Suppose for a contradiction that no such \(\delta\) exists. In particular, for each n=1,2,3… we can find a point \(x_n \in X\) s.t. \(\forall G \in \mathcal{C} \; B_{1/n}(x_n) \not \subset G\). By sequential compactness, we have a convergent subsequence, say \(x_{n_j} \to x\). Then there is some \(G \in C\) with \(x \in G\) and then some \(\epsilon > 0\) with \(B_{\epsilon}(x) \subset G\). Now choose j sufficiently large that both \(x_{n_j} \in B_{\epsilon / 2}(x)\) and also \(1/ n_j < \epsilon / 2\). Then \(B_{1 / n_j}(x_{n_j}) \subset B_{\epsilon}(x) \subset G \in \mathcal{C}\).  Hence \(\delta > 0\) with the desired property does exist.
Now, by previous results, as X is sequentially compact we know that X is totally bounded. So we can find finitely many points \(a_1 , \dots , a_n \in X\) s.t. \(\forall x \in X \; \exists i \in \{1, \dots n\} \; d(x, a_i) < \delta\). In other words, \[X \bigcup_{i=1}^n B_{\delta}(a_i)\].
By definition of \(\delta\), for each i we can choose a set \(G_i \in \mathcal{C}\) s.t. \(B_{\delta}(a_i) \subset G_i\). Let \(\mathcal{B} = \{G_i | 1 \leq i \leq n\}\)
Then \(B \subset \mathcal{C}\), B is finite and \(\cup_{G \in B} G = \cup_{i=1}^n G_i \) which contains \(\cup_{i=1}^n B_{\delta}(a_i) = X\).
So \(\cup_{G \in B} G = X\). So B is a finite subcover of X. Hence X is compact.

Q: Proposition. Let \((X,d)\) be a metric space, \(G \subset X\) and \(F = X \setminus G\). Then G open if and only if F closed.
A: Suppose F is not closed. THen there is a sequence \((x_n)\) in F with \(x_n \to x\) in G. Then for any neighbourhood of x, say \(\mathcal{N}\), \(\exists N \; \forall n \geq N \; x_n \in \mathcal{N}\). But \(\forall n \; x_n \not \in G\). So G is not a neighbourgood of x. But \(x \in G\) so G not open.
Suppose G not open. Then can find \(x \in G\) s.t. \(\forall \epsilon > 0 , B_{\epsilon}(x_n) \not \subset G\). For n = 1,2,3.. we can thus pick \(x_n \in B_{\epsilon}(x) \cap F\) where \(\epsilon = 1/n\). Thus \((x_n)\) is a sequence in F with \(x_n \to x \not \in F\). So F not closed.

Q: Proposition. Let \((X,d)\) be a metric space and let \(\tau = \{G \subset X | G \text{ open}\}\). Then
i) \(\emptyset \in \tau, X \in \tau\)
ii)If \(\sigma \subset \tau\) with \(\sigma \neq \emptyset\) then \(\bigcup_{G \in \sigma} G \subset \tau\)
iii) If \(G_1, \dots G_n \in \tau\) then \(\bigcap_{i=1}^n G_i \in \tau\)
A: i) trivial
ii) Let \(H = \bigcup_{G \in \sigma} G\) and \(a \in H\). Then \(\exists G \in \sigma\) such that \(a \in G\). Then \(a \in G \subset H\) with G open so H is a neighbourgood of a. But \(a \in H\) arbitrary so H is open.
iii) Let \(J \bigcap_{i=1}^n G_i\) and \(a \in J\). For each i, \(a \in G_i\) and \(G_i\) open so can find \(\delta_i > 0\) such that \(B_{\delta_i}(a) \subset G_i\). Then let \(\delta = \min_i \delta_i\). Then \(\delta > 0\) and \(B_{\delta}(a) \subset J\). So J is open.

Q: Defintion. A topological space is a set X endowed with a topology \(\tau\). That is \(\tau \subset \mathcal{P}(X)\) with
[...]
A: i) \(\emptyset, X \in \tau\)
ii) If \(\sigma \subset \tau\) with \(\sigma \neq \emptyset\), then \(\bigcup_{G \in \sigma} G \in \tau\)
iii) If \(G_1 , \dots , G_n \in \tau\) then \(\bigcap_{i=1}^n G_i \in \tau\)
Back Extra: iii) is equivalent to \(G,H \in \tau \implies G \cap H \in \tau\)
Often write just \((X, \tau)\) is a topological space or just X is a topological space.

Definition. Let \((X, \tau)\) be a toplogical space. A set \(G \subset X\) is open iff \(G \in \tau\) and G is closed if \(X \setminus G \in \tau\). A set \(\mathcal{N} \subset X\) is a neighbourgood of \(a \in X\) iff there is an open \(G \subset X\) with \(a \in G \subset \mathcal{N}\).
Let \((Y, \sigma)\) be a toplogical space and \(f : X \to Y\). Then f is continuous if whenever \(G \subset Y\) is open then \(f^{-1}(G) \subset X\) is open, i.e. \(G \in \sigma \implies f^{-1}(G) \in \tau\)
We say f is continous at a if whenever \(N \subset Y\) is a neighbourgood of f(a) then \(f^{-1}(\mathcal{N}) \subset X\) is a neighbourhood of a. We say f is a homeomorphism and that X,Y are homeomorphic if f is a continouous function with continuous inverse.
A topological property is something preserved by homeomorphism.

Q: Proposition. Let \(X,Y\) be toplological spaces and \(f : X \to Y\). Then f is continouous iff f is continous at every \(x \in X\).
A: Let \(a \in X\) and \(\mathcal{N} \subset Y\) be a neighbourhood f(a). Then \(\exists G \subset Y\) open s.t. \(f(a) \in G \subset \mathcal{N}\). As f continuous, \(f^{-1}(G) \subset X\) open and \(a \in f^{-1}(G) \subset f^{-1}(\mathcal{N})\). So \(f^{-1}(\mathcal{N})\) is a neighbourhood of a.
Let \(G \subset Y\) be open. Let \(a \in f^{-1}(G)\). Then \(f(a) \in G\) and G open so G is a neighbourhood of \(f(a)\). As f continuous at a, \(f^{-1}(G)\) is a neighbourgood of a. So \(f^{-1}(G)\) is a neighbourhood of all of its points, so \(f^{-1}(G)\) open. So f is continuous.

Q: Proposition. Let \((X, \tau), (Y, \sigma), (Z, \rho)\) be toplogical spaces and \(f: X \to Y\), \(g: Y \to Z\) continuous. Then \(g \circ f : X \to Z\) is continouous.
A: Let \(G \in \rho\). As g continuous, \(g^{-1}(G) \in \sigma\). As f is continuous, \(f^{-1}(g^{-1}(G)) \in \tau\) so \((g \circ f)^{-1}(G) \in \tau\)

Q: Definition. Let X be a topological space, \((x_n)\) a sequence in X and \(x\in X\). We say \((x_n)\) converges to x and write \(x_n \to x\) if [...]
A: whenever \(\mathcal{N}\) is a neighbourhood of x in X then \[\exists N \; \forall n \geq N \; x_n \in \mathcal{N}\]

Q: Definition. A topological space \((X, \tau)\) is Hausdorff if [...]
A: whenever \(x,y \in X\) with \(x \neq y\), then there are \(G,H \in \tau\) with \(x \in G, y \in H\) and \(G \cap H = \emptyset\).

Q: Proposition. Let X be a Hausdorff topological space and \((x_n)\) be a sequence in X with \(x_n \to x\) and \(x_n \to y\). Then \(x = y\).
A: Suppose \(x \neq y\). Then we can find disjoint, open \(G,H \subset X\) with \(x \in G\) and \(y \in H\). Then G is a neighbourhood of x and \(x_n \to x\) so \(\exists N_1\) s.t. \(\forall n \geq N_1, x_n \in G\). Similarly \(\exists N_2 \text{ s.t } x_n \in H \; \; \forall n \geq N_2\) . But let \(N= \max (N_1 , N_2)\). Then \(x_N \in G \cap H\). Contradiction.

Q: Proposition. Let X,Y be topological spaces, let \(f:X \to Y\), let \(x \in X\) and suppose f is continuous at x. Let \((x_n)\) be a sequence in X with \(x_n \to x\). Then \(f(x_n) \to f(x)\).
A: Let \(N\) be a neighbourhood of f(x) in Y. So \(f^{-1}(N)\) is a neighbourhood of x in X as f is continuous at x. As \(x_n \to x\), we have some M s.t. \(\forall n \geq M\), \(x_n \in f^{-1}(N)\). Hence \(\forall n \geq M , f(x_n) \in N\). So \(f(x_n) \to f(x)\).

Q: Definition. Let \((X, \tau)\) be a topological space with \(Y \subset X\). The subspace topology on Y is [...]
A: \[\{G \cap Y | G \in \tau\}\]

Q: Proposition. Let (X,d) be a metric space and let Y be a metric subspace of X. Then the topology induced by the metric Y inherits from X is exactly the subspace topology on Y thought of as a subspace of \((X , \tau)\) where \(\tau\) is the topology induced by d on X.
A: Let \(\sigma\) be the toplogy on Y induced by \(d |_{Y^2}\). We'll write \(B_{\delta}(x)\) for open balls in the metric space X and \(A_{\delta}(x)\) for open balls in the metric space Y. Note for any \(y \in Y\) and \(\delta > 0\), \(X_{\delta}(y) = B_{\delta}(y) \cap Y\).
First suppose \(G \in \tau\). Let \(y \in G \cap Y\). Then \(y \in G\) and \(G \subset X\) open so there is some \(\delta > 0\) with \(B_{\delta}(y) \subset G\). But then \(A_{\delta}(y) \subset G \cap Y\). So \(G \cap Y \in \sigma\).
Suppose instead that \(H \in \sigma\). Then for each \(y \in H\) we can find \(\delta_y > 0\) s.t. \(A_{\delta_y}(y) \subset H\). Hence \[H = \bigcup_{y \in H} A_{\delta_y}(y)\] Now let \(G = \bigcup_{y \in H} B_{\delta_y}(y)\). Then \(G \in \tau\). Moreover, \[G \cap Y = \bigcup_{y \in H}(B_{\delta_y}(y) \cap Y) = \bigcup_{y\in G} A_{\delta_y}(y) = H\]

Q: Proposition. A subspace of a Hausdorff space is Hausdorff.
A: Let \((X, \tau)\) be a Hausdorff space, let Y be a subspace of X and let \(x,y \in Y\) with \(x \neq y\). As X Hausdorff, we have disjoint, open \(G,H \subset X\) with \(x \in G\) and \(y \in H\). Then \(G \cap Y, H \cap Y\) are disjoint open subsets of Y with \(x \in G \cap Y\) and \(y \in G \cap Y\). So Y is Hausdorff.

Definition. Let \((X, \tau)\) be a topological space. An open cover of X is a collection \(\mathcal{C} \subset \tau\) s.t. \(X = \cup_{G \in \mathcal{C} } G\) . A subcover of \(\mathcal{C}\) is a subset \(\mathcal{D} \subset \mathcal{C}\) s.t. \(\mathcal{D}\) is an open cover of X. We say X is compact if every open cover has a finite subcover. We say X is sequentially compact if every sequence in X has a convergent subsequence.

Q: Proposition. A continuous real-valued function on a sequentially compact topological space is bounded and attains its bounds.
A: As usual we assume \(\mathbb{R}\) has the usual metric/topology unless otherwise stated.
The proof is the same as for metric spaces.

Q: Proposition. A continuous image of a compact space is compact.
A: Let \(f: X \to Y\) be continuous with X compact. Let \(\mathcal{C}\) be an open cover of \(f(X)\) in Y. Then \(\{f^{-1}(G) | G \in \mathcal{C}\}\) is an open cover of X so it has a finite subcover, say \(\{f^{-1}(G_1), \dots , f^{-1}(G_n)\}\). Then \(\{G_1 , \dots , G_n\} \in \mathcal{C}\) is an open cover of f(X) in Y.

Q: Theorem. A continuous, real-valued function on a compact topological space is bounded and attains its bounds.
A: Let \(f : X \to \mathbb{R}\) be continuous with X compact and the usual topology on \(\mathbb{R}\). So f(X) is compact, i.e. f(X) is closed and bounded. In particular, f is bounded.
Let \(\sigma \sup f(X)\) and \(\tau = \inf f(X)\). By definition of \(\sigma\), there is a sequence \((x_n)\) in f(X) with \(x_n \to \sigma\). But f(X) closed so \(\sigma \in f(X)\). Similarly \(\tau \in f(X)\).

Q: Definition. Let \((X, \tau)\) be a topological space and let \(a,b \in X\). A path from a to b in X is [...] We say X is path-connected if [...]
A: a continuous function \(\psi : [0,1] \to X\) s.t \(\psi(0) = a\) and \(\psi(1) = b\).; whenever \(a,b \in X\) there is a path in X from a to b.

Q: Definition. Let X be a topological space. We say X is disconnected if [...] If X is not disconnected, we say X is connected.
A: there are non-empty, open \(U,V \subset X\) with \(X = U \cup V\) and \(U \cap V = \emptyset\). When this happens we say the sets U,V disconnect X.
Back Extra: Let \(S \subset X\). Then S disconnected iff there exist \(U,V \subset X\) open s.t. \(S \subset U \cup V\) and \(S \cap U \cap V = \emptyset\) but \(S \cap U \neq \emptyset\) and \(S \cap V \neq \emptyset\). In this case we'll say U,V disconnect S in X.
U and V don't intersect in S, but do intersect in X.

Q: Proposition. Let X be a topological space. Then X is connected iff every continuous \(f: X \to \mathbb{Z}\) with usual topology is constant.
A: Suppose X connected and \(f:X \to \mathbb{Z}\) is continuous. Then for each \(n \in \mathbb{Z}\) we have \(\{n\}\) is both closed and open, so \(f^{-1}(\{n\})\) is open and closed, so it is either the empty set or X. So f is constant.
Suppose X is not connected. Let U,V disconnect X. Define \(f: X \to \mathbb{Z}\) by f(x) = 1 if \(x \in U\) and 0 if \(x \in V\). \(U,V \neq \emptyset\) so f is not constant. And if \(Y \subset Z\) then \(f^{-1}(Y)\) is one of \(\emptyset, U,V,X\) and so \(f^{-1}(Y)\) is open. So f continuous.

Q: Proposition. The continuous image of a connected space is connected.
A: Let \(f: X \to Y\) be continuous and let X be connected. Suppose \(U,V \subset Y\) disconnect \(f(X)\). Then \(f^{-1}(U), f^{-1}(V)\) disconnect X. Contradiction.

Q: Proposition. Let \(I \subseteq \mathbb{R}\) with usual topology. Then I connected iff I is an interval.
A: Suppose I is not an interval. Then can find a,b,c with \(a,b \in I, c \not \in I\) and \( a < c < b\). So \((-\infty, c)\) and \((c, \infty)\) disconnect I in \(\mathbb{R}\).
Suppose I is an interval. Then by IVT any continuous \(f: I \to \mathbb{Z}\) is constant, so by our proposition this means I is connected.

Q: Proposition. Any path-connected space is connected.
A: Let X be path-connected and suppose U,V disconnect X. Pick \(u \in U, v \in V\), and let \(\psi\) be a path from u to v. Then U,V disconnect \(\psi([0,1])\) in X. But the continuous image of a connected space has to be connected. Contradiction.

Q: Definition. Let X be a topological space. For \(x,y \in X\) write \(x \to y\) if [...]
A: there is a path in X from x to y.
Back Extra: This is an equivalence relation on X. The equivalence classes are called the path-components of X.

Q: Theorem. An open, connected subset of \(\mathbb{R}^n\) with Euclidean topology is path-connected.
A: Let \(X \subset \mathbb{R}^n\) be open and connected. If \(X = \emptyset\) we are done, so assume \(X \neq \emptyset\) and fix \(a \in X\). Let \(U = [a]_{\rightarrow}\). First, \(a \to a\) so \(a \in U\) so \(U \neq \emptyset\). Let \(b \in X\). As X open we have some \(\delta > 0\) s.t. \(B_{\delta}(b) \subset X\). Note \(\forall c \in B_{\delta}(b)\) then \(b \to c\). Suppose \(b \in U\). Then \(a \to b\), and \(\forall c \in B_{\delta}(b), b \to c\) so \(\forall c \in B_{\delta}(b), a \to c\) so \(B_{\delta}(b) \subset U\). So U is open.
Suppose \(b \in X \setminus U\). Let \(c \in B_{\delta}(b)\) and suppose \(a \to c\). Then, as \(c \to b\) we have \(a \to b\) and so \(b \in U\) which is a contradiction. Hence \(B_{\delta}(b) \subset X \setminus U\). Hence \(X \setminus U\) is open and so U is closed.
Now U is open, closed and non-empty so by connectedness U = X. Hence X is path-connected.

Q: Proposition. Let X be a topological space. For \(x,y \in X\) write \(x \sim y\) if \(\exists A \subset X\) connected with \(x,y \in A\). Then \(\sim\) is an equivalent relation on X. The equivalence classes of X are precisely the maximal connected subset of X and each connected subset of X is contained in some equivalence class.
A: Let \(x,y,z \in X\). We have \(x \sim x\)(as {x} connected) and clearly \(x \sim y \implies y \sim x\) as the definition is symmetric.
Suppose \(x \sim y\) and \(y \sim z\). Let \(A,B \subset X\) connected with \(x,y \in A\) and \(y,z \in B\). Then \(x,z \in A \cup B\). Suppose \(U,V \in X\) disconnect \(A \cup B\), wlog \(y \in U\)and \(A \cap V \neq \emptyset\). But \(y \in A\) so \(U \cap A \neq \emptyset\) so \(U,V\) disconnect A. So \(A \cup B\) connected and so \(x \sim z\).
Let \(x \in X\) and suppose \(U,V\) disconnect \([x]_{\sim}\), wlog \(x \in U\). Let \(y \in V \cap [x]_{\sim}\). As \(x \sim y\) be have a connected \(A \subset X\) s.t. \(x,y \in A\). Now \(A \subset [x]_{\sim}\) and \(x \in A\cap U, y \in A \cap V\) so \(U,V\) disconnect A. Contradiction. So \([x]_{\sim}\) is connected.
Suppose \([x]_{\sim} \subset B\) with B connected. If \(y \in B\) then \(x,y \in B\) so \(x \sim y\) so \(y \in [x]_{\sim}\). So \(B \subset [x]_{\sim}\). So \([x]_{\sim} = B\).
Let \(A \subset X\) be connected, wlog \(A \neq \emptyset\). Pick \(x \in A\). Then \(A \subset [x]_{\sim}\).
Finally suppose A is maximal connected. Then \(A \subset [x]_{\sim}\) with \([x]_{\sim}\) connected so \(A = [x]_{\sim}\)

Q: Definition. The (connected) components of X are [...]
A: the equivalence classes of \(\sim\).
Back Extra: \(x \sim y\) if \(\exists A \subset X\) connected with \(x,y \in A\).

Q: Proposition. Let \((X, \tau)\) and \((Y , \sigma)\) be topological spaces and let \(\pi\) be the product topology on \(X \times Y\). Then
i) X,Y Hausdorff \(\implies X \times Y\) Hausdorff
ii) X,Y compact \(\implies X \times Y\) compact
iii)X,Y connected \(\implies X \times Y\) connected
A: i) Let \(a,b \in X \times Y\) with \(a \neq b\). Then WLOG \(a = (x,y)\) and \(b = (z,w)\) with \(x \neq z\). As X Hausdorff there are \(U,V \in \tau\) with \(x \in U, z \in V\) and \(U \cap V = \emptyset\).
Then \(U \times Y, V \times Y \in \pi\)  with \(a \in U \times Y\), \(b \in V \times Y\) and \((U \times Y) \cap (V \times Y) = \emptyset\).
ii) Let \(\mathcal{C}\) be an open cover of \(X \times Y\). Each \(E \in \mathcal{C}\) is a union of sets from \(G \times H\) with \(G \in \tau\) and \(H \in \sigma\); replace each \(E \in \mathcal{C}\) by all the sets in this union to obtain a new open cover \[\mathcal{C}' = \{G_i \times H_i | i \in I\}\] where I is some index set and \(\forall i \in I, G_i \in \tau\) and \(H_i \in \sigma\).
For each \(x \in X\) do the following: let \(J_x = \{i \in I | x \in G_i\}\). Then \(\{H_i | i \in J_x\}\) is an open cover of Y so, by compactness of Y, it has a finite subcover, say \(\{H_{i^{x}} , \dots , H_{i^{(x)}_{m_x}}\}\). Now let \(G^{(x)} = \bigcap_{\alpha = 1}^{m_x} G_{i_{\alpha}}(x)\). Then \[G^{(x)} \times Y \subset \bigcup_{\alpha = 1}^m (G_{i^{(x)}_{\alpha}} \times H_{i_{\alpha}^{(x)}})\] Now \( \{G^{(x)} | x \in X\}\) is an open cover of X, so by compactness of X has a finite subcover, say \[\{G^{(x_1)}, \dots , G^{(x_n)}\}\]. Then \(\{G_{i_{\alpha}^{x_j}} \times H_{i_{\alpha}^{(x_j)}} | 1 \leq j \leq n, 1 \leq \alpha \leq m_{x_j}\}\) is a finite subcover of \(\mathcal{C}'\). But each \(D \in \mathcal{C}'\) is connected is some set in \(\mathcal{C}\). So replacing each of the sets in the thing above by such a set in \(\mathcal{C}\) containing it we get a finite subcover of \(\mathcal{C}\).
iii) For each \(x \in X\), the space \(\{x\} \times Y\) is homeomorphic to Y. So \(\{x\} \times Y\) is connected and lies entirely within a single component of \(X \times Y\). Similarly for each \(y \in Y, X \times \{y\}\) lies entirely within a single component of \(X \times Y\). But components partition \(X \times Y\) so there is a only one component so \(X \times Y\) is connected.

Q: Definition. The product topology on \(X \times Y\) is the topology such that a subset U of \(X \times Y\) is open if [...]
A: there exists a set I and open sets \(U_i ,V_i\), in \(X,Y\) for all \(i \in I\) such that \[U = \bigcup_{i \in I} U_i \times V_i\]

Q: Definition. Let \((X, \tau)\) be a topological space and \(\sim\) an equivalence relation on X. Let \(q : X \to X / \sim = \{[x]_{\sim} | x \in X\}\) be the quotient map, \(q(x) = [x]_{\sim}\). The quotient topology on \(X/\sim\) is defined by [...]
A: \[\kappa = \{G \subset X / \sim | q^{-1}(G) \in \tau\}\]
Back Extra: It is clearly a topology, and the largest topology on \(X / \sim\) making q continuous.
Compactness and connectedness preserved. Hausdorff not so \(X \text{ Hausdorff} \not \implies X/\sim \text{ Hausdorff}\). Take \(\mathbb{R}\)with the usual topology, let \(x \sim y \iff x-y \in \mathbb{Q}\). Suppose G is open in \(\mathbb{R} / \sim\). Then \(q^{-1}(G) \in \mathbb{R}\) so , if \(G \neq \emptyset\), G contains some non-trivial open interval I. So \(q(I) = \mathbb{R} / \sim\). So \(G = \mathbb{R} / \sim\). So quotient topology on \(\mathbb{R} / \sim\) is the indiscrete topology which isn't Hausdorff.

Q: Proposition. Let \((X, \tau)\) be a topological space and \(\sim\) an equivalence relation on X. Let \(q : X \to X / \sim\) be quotient map. Let \(f: X \to Y\) be a continuous function respecting \(\sim\) so \(x \sim y \implies f(x) = f(y)\). Then there is a unique \(\overline{f}\) continuous, \(\overline{f}: X / \sim \to Y\) s.t. \(f = \overline{f} \circ q\).
A: For \(x \in X\) let \(\overline{f}([x]_{\sim}) = f(x)\).
Well defined: \([x]_{\sim} = [y]_{\sim} \implies x \sim y \implies f(x) = f(y)\)
Continuous: If \(G \subset Y\) open then \(q^{-1}(\overline{f}^{-1}(G)) = (\overline{f} \circ q)^{-1}(G) = f^{-1}(G)\) is open in X, so \(\overline{f}^{-1}(G)\) open in \(X / \sim\).
Also, \(\forall x \in X \; \overline{f}(q(x)) = \overline{f}([x]_{\sim}) = f(x)\).
Suppose also \(g: X / \sim \to Y\) works. Then \(\forall x \in X, g([x]_{\sim}) = g(q(x)) = f(x) = \overline{f}([x]_{\sim})\). So \(g = \overline{f}\).

Q: Theorem. Topological Inverse Function Theorem.
(i)A closed subset of a compact space is compact
(ii)A compact subset of a Hausdorff space is closed
(iii)A continuous bijection from a compact space to a Hausdorff space is a homomorphism
A: (i)Let X be compact and \(F \subset X\) be closed. Let \(\mathcal{C}\) be an open cover of F. Then \(\mathcal{C} \cup \{X / F\}\) is an open of X so has a finite subcover \(\mathcal{D}\). So \(\mathcal{D} / \{X / F\}\) is a finite subcover of \(\mathcal{C}\).
(ii) Let X be Hausdorff and \(K \subset X\) be compact. Let \(y \in X / K\). For each \(x \in K\) there are disjoint, open \(U_x, V_x \subset X\) s.t. \(x \in U_x, y \in V_x\). Then \(\{U_x | x \in K\}\) is an open cover of K so has a finite subcover, say \(\{U_{x_1}, \dots , U_{x_n}\}\). Let \(U = \bigcup_{i=1}^n U_{x_i}\) and \(V = \bigcap_{i=1}^n V_{x_i}\). Then \(K \subset U, U \cap V = \emptyset, V\) open and \(y \in V\). So have \(y \in V \subset X / K\) with V open so \(X / K\) is a neighbourhood of y. But \(y \in X / K\) was arbitrary. So \(X / K\) open. So K closed.
(iii) Let \(f : X \to Y\) be a continuous bijection with X compact and Y Hausdorff. Let \(G \subset X\) be open. As f bijective, \[(f^{-1})^{-1}(G) = f(G) = Y \setminus f(X \setminus G)\] Now G open so \(X \setminus G \) closed so \(X \setminus G\) compact by (i), so \(f(X \setminus G)\) is compact (as f continuous) so \(f(X \setminus G)\) closed in Y by (ii) so \(Y \setminus f(X \setminus G)\) open, i.e. \((f^{-1})^{-1}(G)\) open. So \(f^{-1}\) continuous. Hence f is a homeomorphism.

Q: Definition. Let \(f : \mathbb{R}^n \to \mathbb{R}^m\). We say f is differentiable at \(x \in \mathbb{R}^n\) if [...]
A: there is a linear map \(\alpha : \mathbb{R}^n \to \mathbb{R}^m\) with \[f(x+h) = f(x) + \alpha(h) + \epsilon(h)||h||\]where \(\epsilon(h) \to 0\) as \(h \to 0\)

Q: Proposition. Let \(f: \mathbb{R}^n \to \mathbb{R}^m, x \in \mathbb{R}^n,  \alpha, \beta \in L(\mathbb{R}^n, \mathbb{R}^m)\) s.t. \[f(x+h) = f(x) + \alpha(h) + \epsilon(h)\] \[f(x+h) = f(x) + \beta(h) + \eta(h)\]
where \(\epsilon(h), \eta(h) \to 0\) as \(h \to 0\). Then \(\alpha = \beta\)
A: For any \(h \in \mathbb{R}^n\) and \(\lambda \in \mathbb{R} \setminus \{0\}\), we have \[\alpha(h) - \beta(h) = (\eta(h) - \epsilon(h)) || h ||\] and so \[|| \alpha(h) - \beta(h)|| = ||1/\lambda (\alpha(\lambda h) - \beta(\lambda h))|| = || 1/\lambda (\eta (\lambda h) - \epsilon (\lambda h) )||\lambda h |||| = || \eta(\lambda h) - \epsilon(\lambda h)|| ||h|| \to 0\] as \(\lambda \to 0\). So \(\alpha(h) = \beta(h)\)

Q: Definition. If \(f: \mathbb{R}^n \to \mathbb{R}^m\) is differentiable at \(x \in \mathbb{R}^n\), we can define the derivative of f at x to be the unique linear map \(D f|_x : \mathbb{R}^n \to \mathbb{R}^m\) s.t. [...]
A: \[f(x+h) = f(x) + D f|_x (h) + \epsilon(h) ||h||\] with \(\epsilon(h) \to 0\) as \(h \to 0\)

Q: Proposition. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) be differentiable at x. Then f is continuous at x.
A: Write \(f(x+h) = f(x) + \alpha(h) + \epsilon(h) ||h||\) where \(\alpha \in L (\mathbb{R}^n, \mathbb{R}^m)\) and \(\epsilon(h) \to 0\) as\(h \to 0\). Then \(\alpha\) continuous so \(\alpha(h) \to \alpha(0) = 0\) as \(h \to 0\) so \(f(x+h) \to f(x) + 0 + 0 \times 0 = f(x)\) as \(h \to 0\).

Q: Definition. For \(\alpha \in L(\mathbb{R}^n, \mathbb{R}^m)\), we define the operator norm of \(\alpha\) to be [...]
A: \[||\alpha || = \sup \{||\alpha x|| \bigg | x \in \mathbb{R}^n, ||x|| = 1\}\]
Back Extra: This is well-defined because \(\alpha\) is continuous and \(\{x \in \mathbb{R}^n | ||x|| = 1\}\) is compact. So we are bounded.

Q: Proposition. Let \(\alpha, \beta \in L(\mathbb{R}^n, \mathbb{R}^m), \lambda \in \mathbb{R}\) and \(x \in \mathbb{R}^n\). Then:
(i) \(||\alpha|| \geq 0\) with equality iff \(\alpha = 0\);
(ii) \(||\lambda \alpha || = |\lambda| \; ||\alpha||\);
(iii) \(|| \alpha + \beta|| \leq || \alpha || + || \beta ||\);
(iv) \(||\alpha x || \leq || \alpha || \; ||x||\); and
(v) \(|| \alpha \beta || \leq || \alpha || \; || \beta||\)
A: (i), (ii) are obvious.
(iii) Let \(y \in \mathbb{R}^n\), ||y|| = 1. Then \[||(\alpha + \beta)(y)|| \leq || \alpha y|| + || \beta y|| \leq ||\alpha|| + || \beta||\]
(iv) WLOG \(x \neq 0\). Then \[ || \alpha x || = || \alpha (\frac{x}{||x||}) ||x|| \; || \leq || \alpha (\frac{x}{||x||}) || \; ||x|| \leq ||\alpha || \; ||x ||\]
(v) Let \(y \in \mathbb{R}^n\) , \(||y|| = 1\). Then \( || \alpha \beta(y)|| \leq ||\alpha || ||\beta y || \leq ||\alpha || \; || \beta|| \) so \(|| \alpha \beta || \leq ||\alpha || \; || \beta ||\)

Q: Proposition. Let \(f,g : \mathbb{R}^n \to \mathbb{R}^m\) and let \(\lambda : \mathbb{R}^n \to \mathbb{R}\) all be differentiable at \(x \in \mathbb{R}^n\). Then the point wise sum and product \(f+g, \lambda f : \mathbb{R}^n \to \mathbb{R}^m\) are differentiable at x with \[D(f+g)|_x = D f|_x + D g|_x\] and \[D(\lambda f)|_x (h) = \lambda (x) D f|x(h) + D \lambda |_x (h) f(x)\]
A: Write \(f(x+h) = f(x) + \alpha (h) + \epsilon (h) ||h||\), \(g(x+h) = g(x) + \beta (h) + \eta (h) ||h||\) and \(\lambda (x+h) = \lambda (x) + \gamma(h) + \xi (h) ||h||\)
where the maps have the properties we'd expect. Then \[(f+g)(x+h) = (f+g)(x) + (\alpha + \beta)(h) + (\epsilon(h) + \eta(h))||h||\] where \(\alpha + \beta\) linear and \(\epsilon(h) + \eta (h) \to 0\) as \(h \to 0\). Also \[(\lambda f)(x+h) + (\lambda f)(x) + \lambda (x)\alpha (h) + \gamma (h) f(x) + \zeta(h)||h||\] where the second term is linear in h and the last is an error term where \[\zeta(h) + \lambda(x)\epsilon(h) + \frac{\gamma(h) \alpha(h)}{||h||} + \gamma(h) \epsilon(h) + \xi(h)f(x+h) \to 0\] as \(h \to 0\) since \(\lambda(x) \epsilon(h) \to \lambda(x) \times 0 = 0\);
\[\frac{\gamma(h) \alpha(h)}{||h||} = || \gamma(h) \alpha (\frac{h}{||h||}) || \leq ||\alpha|| ||\gamma(h)|| \to 0\];
\[\gamma(h)\epsilon(h) \to \gamma(0) \times 0 = 0\] and \[\xi(h)f(x+h) \to 0 \times f(x) = 0\] as f continuous.

Q: Theorem. Chain Rule. Let \(f: \mathbb{R}^p \to \mathbb{R}^n\) be differentiable at \(x \in \mathbb{R}^p\) and \(g: \mathbb{R}^n \to \mathbb{R}^m\) be differentiable at \(f(x) \in \mathbb{R}^n\). Then \(g \circ f\) is differentiable at x and \[D(g \circ f) |_x = D g|_{f(x)} \circ D f|_x\]
A: Write \(f(x+h) = f(x) + \alpha(h) + \epsilon(h) || h||\) and \(g(f(x)+k) = g(f(x)) + \beta(k) + \eta (k)||k||\) where the maps have the expected properties. Then \[g(f(x+h)) = g(f(x) + \alpha(h) + \epsilon(h) ||h||) = g(f(x)) + \beta(\alpha(h) + \epsilon(h) || h||) + \eta(\alpha(h) + \epsilon(h)||h||)|| \alpha(h) + \epsilon(h)||h|| \; || = g(f(x)) + (\beta \circ \alpha)(h) + \zeta(h)||h||\] where \[\zeta(h) = \beta(\epsilon(h)) + \eta (\alpha(h) + \epsilon(h) ||h||) || \frac{\alpha(h)}{||h||} + \epsilon(h)||\] Assume wlog \(\eta(0) = 0\) so \(\eta\) continuous at 0. Then \[\beta(\epsilon(h)) \to \beta(0) = 0\] as \(h \to 0\);
\[\alpha(h) + \epsilon(h)||h|| \to 0 + 0 \times 0\] as \(h \to 0\) so \(\eta(\alpha(h) + \epsilon(h) || h||) \to 0\) and \[|| \frac{\alpha(h)}{||h||} + \epsilon(h)|| = || \alpha (\frac{h}{||h||}) + \epsilon(h) || \leq ||\alpha || + || \epsilon(h)|| \to ||\alpha ||\] as \(h \to 0\). Hence \(\zeta(h) \to 0\) as \(h \to 0\). So we are done.

Q: Proposition. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) and \(x \in \mathbb{R}^n\). Then f is differentiable at x iff for each \(i = 1, \dots , m\) \(f_i\) is differentiable at x where \(f_i : \mathbb{R}^n \to \mathbb{R}\) s.t. \[f = (f_1 , \dots , f_m)^T\]
More over, if f is differentiable then for each i, \(D f_i |_x = (Df|_x)_i\) and the matrix A of \(D f|_x\) w.r.t the standard bases of \(\mathbb{R}^n, \mathbb{R}^m\) is given by \(A_{ij} = D_j f_i(x)\)
A: Write \(f(x+h)\ = f(x) + \alpha(h) + \epsilon(h) ||h||\) where \(\alpha \in L(\mathbb{R}^n, \mathbb{R}^m)\). Then for each i, \[f_i(x+h) = f_i(x) + \alpha_i(h) + \epsilon_i(h)||h||\] with \(\alpha_i \in L(\mathbb{R}^n, \mathbb{R})\), and \(\epsilon(h) \to 0\) iff for each i, \(\epsilon_i(h) \to 0\) as \(h \to 0\).
Finally, assume f differentiable at x and has \(Df|_x = \alpha\), and let \(e_1 , \dots , e_n\) be standard basis of \(\mathbb{R}^n\) and \(e_1', \dots e_m'\) the standard basis of \(\mathbb{R}^m\). Then \[\frac{f(x+te_j ) - f(x)}{t} = \frac{\alpha (te_j) + \epsilon(t e_j)|| t e_j ||}{t} = \alpha(e_j) \pm \epsilon(te_j) \to \alpha(e_j)\] as \(t \to 0\). That is \(D_jf(x) = \alpha(e_j)\), so \(\alpha(e_j) = \sum_{i=1}^m D_j f_i(x) e_i'\).

Q: Proposition. Let \(m,n \geq 1\) and let \(|| . ||, ||.||'\) be the operator and Euclidiean norms on \(M_{m \times n}\) respectively. Then there are constant c,d s.t. \(\forall A \in M_{m \times n}\), \[||A|| \leq c ||A||' \text{ and } ||A||' \leq d||A||\]
A: Let \(x \in \mathbb{R}^n\) with \(||x|| = 1\). Then \[||Ax||^2 = \sum_{j=1}^m ((Ax)_j)^2 = \sum_{j=1}^m (\sum_{i=1}^n A_{ji}x_i)^2 \leq \sum_{j=1}^m(\sum_{i=1}^n ||A||')^2 = mn^2 ||A||'^2\] so \[||A|| \leq n \sqrt{m} ||A||'\]
Also, \[||A||'^2 = \sum_{i=1}^n \sum_{j=1}^n A_{ij}^2 = \sum_{j=1}^n ||Ae_j||^2 \leq \sum_{j=1}^n ||A||^2 = n||A||^2\] so \[||A||' \leq \sqrt{n}||A||\]

Q: Theorem. Mean Value Inequality. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) and let \(a,b \in \mathbb{R}^n\). Suppose f is differentiable at z for all \(z \in [a,b] = \{a + t(b-a) | t \in [0,1]\}\). Then \[||f(b) - f(a)|| \leq ||b-a|| \sup_{z \in [a,b]} ||Df|_z||\]
A: Define \(F : [0,1] \to \mathbb{R}\) by \[F(t) = f(a + t(b-a)) \cdot (f(b) - f(a))\] So \(F = \alpha \circ f \circ \phi\) where \(\phi : [0,1] \to \mathbb{R}^n\) with \(\phi(t) = a+ t(b-a)\) and \(\alpha : \mathbb{R}^m \to \mathbb{R}\) with \(\alpha(y) = y \cdot (f(b) - f(a))\). Clearly F is continuous on [0,1].
Also \(\alpha\) is linear so everywhere differentiable with \(D \alpha|_y = \alpha\), and \(\phi\) is differentiable on (0,1) with \(\phi'(t) = b-a\), i.e. \(D \phi|_t (h) = h(b-a)\). Hence by Chain rule F is differentiable on \((0,1)\) with \[DF|_t(h) = D \alpha|_{f(\phi(t))} (D f|_{\phi(t)} (D \phi |_t (h))) = (f(b) - f(a)) \cdot D f |_{\phi(t)} (h(b-a)) = h(f(b) - f(a)) \cdot D f|_{\phi(t)}(b-a)\] That is \(F'(t) = (f(b) - f(a)) \cdot D f|_{\phi(t)}(b-a)\). Now, \(||f(b) - f(a)||^2 = (f(b) - f(a)) \cdot (f(b) - f(a)) = |F(1) - F(0)| = |F'(t)| = |(f(b) - f(a)) \cdot D f|_z (b-a)| \leq ||f(b) - f(a)|| || Df|_z (b-a)||\) by Cauchy-Schwartz. If \(f(a) = f(b)\) we are done, otherwise \[||f(b) - f(a)|| \leq ||D f|_z (b-a)|| \leq ||Df|_z||||b-a||\]

Q: Corollary. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) and let \(G \subset \mathbb{R}^n\) be open and connected. Suppose for all \(z \in G\), f is differentiable at z with \(D f|_z\) being the zero map. Then f is constant on G.
A: Work in subspace topology on G. Let \(G_x = \{u \in G | f(u) = x\}\) for each \(x \in \mathbb{R}^m\). Suppose \(u \in G_x\). Then G open so there is some \(\delta > 0\) s.t. \(B_{\delta}(u) \subset G\). But \(B_{\delta}(u)\) is convex, so if \(a,b \in B_{\delta}(u)\) then \([a,b] \subset B_{\delta}(u)\) so by MVI \(f(b) = f(a)\). SO f constant on \(B_{\delta}(u)\). So \(B_{\delta}(u) \subset G_x\). So \(G_x\) open in G.
But also \(G \setminus G_x = \bigcup_{y \in \mathbb{R}^m \setminus \{x\}} G_y\) is also open in G so \(G_x\) closed in G. But G connected so \(G_x = \emptyset \) or \(G_x = G\). So f constant on G.

Q: Theorem. Let \(f : \mathbb{R}^n \to \mathbb{R}^m\) and \(z \in \mathbb{R}^n\). Suppose all p.d.s of f exists on a neighbourhood of z and are continuous at z. Then f is differentiable at z. How to prove for n=2,m=1?
A: Let \(z = (x,y)\). By MVT, \[f(x+h,y) - f(x,y) = hD_1 f(x+\theta h,y)\] and \[f(x+h, y+k) - f(x+h,y) = kD_2 f(x+h, y + \phi k)\] for some \(\theta, \phi \in (0,1)\) depending on h,k. Nevertheless, \((x+\theta h ,y)\to (x,y)\) and \((x+h, y+k) \to (x,y)\) as \((h,k) \to (0,0)\). Hence \[f(x+h,y+k) - f(x,y) = h(D_1 f(x,y) + o(1)) + k(D_2f(x,y) + o(1)) = hD_1f(x,y) + kD_2(x,y) + o((h,k))\] as \((h,k) \to (0,0)\).
Hence f is differentiable at \((x,y)\) with \(D f|_{(x,y)}(h,k) - hD_1 f(x,y) + h D_2 f(x,y)\)
WLOG m=1 as we can look at each \(f_i\) separately. For n > 2, proof is essentially the same with n applications of MVT.

Q: Theorem. Symmetry of Mixed Partial Derivatives. Let \(f : \mathbb{R}^n \to \mathbb{R}^m\) and \(z \in \mathbb{R}^n\). Let \(i,j \in \{1, \dots , n\}\). Suppose \(D_i D_j f\) and \(D_j D_i f\) exist in a neighbourhood of z and are continuous at z. Then \[D_i D_j f(z) = D_j D_i f(z)\]
A: WLOG n=2, m =1, i = 1, j =2. Write \(z = (x,y)\). For h small, define \(\Delta_h = f(x+h, y+h) + f(x,y) - f(x+h,y) - f(x,y+h)\). So \(\Delta_h = (f(x+h, y+h) - f(x+h,y)) - (f(x,y+h) - f(x,y)) = F_h(x+h) - F_h(x)\) where \(F_h(w) = f(w, y+h) - f(w,y)\). Then \(F_h\) is differentiable on a neighbourhood of x with \(F_h'(w) = D_1 f(w, y+h) - D_1 f(w,y)\). Hence by MVT, \[\Delta_h = h(D_1 f(x+\theta h, y+h) - D_1 f(x+\theta h, y)) = h^2 D_2 D_1 f(x+\theta h, y+ \theta h)\] for some \(\theta, \psi \in (0,1)\) depending on h.
Similarly, \(\Delta_h = h^2 D_1 D_2 f(x+ \zeta h, y + \xi h)\) for some \(\zeta, \xi \in (0,1)\) depending on h. Hence for \(h \neq 0\) small, \[D_1 D_2 f(x+ \zeta h, y + \xi h) = D_2 D_1 f(x + \theta h, y+ \phi h)\] Hence letting \(h \to 0\), by continuity of partials at (x,y) we get \[D_1 D_2 f(x,y) = D_2 D_1 f(x,y)\]

Q: Definition. If \(F(x) = Df|_x\) is differentiable at \(x \in \mathbb{R}^n\) then we say f is twice-differentiable at x with second derivative of f at x is [...]
A: \[D^2 f|_x = DF|_x\]

Q: Definition. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) and let \(x \in \mathbb{R}^n\). We say f is continuously differentiable at x if [...]
A: f is differentiable at every y in some neighbourhood of x and the function \(y \mapsto Df|_y\) is continuous at x.

Q: Theorem. Second-Order Taylor Theorem. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) be twice-differentiable at a. Then \[f(a + h) = f(a) + Df|_a (h) + \frac{1}{2}D^2 f|_a (h,h) + o (||h||^2)\]
A: WLOG m=1(we could consider each \(f_i : \mathbb{R}^n \to \mathbb{R}\) separately).
Now, for each h define \(\phi_h : [0,1] \to \mathbb{R}\) by \(\phi(t) = f(a + th) - f(a) - tDf|_a(h) - \frac{1}{2}t^2 D^2 f|_a (h,h)\).
\(\phi(0) = 0, \phi(1) = f(a+h) - f(a) - Df|_a(h) - \frac{1}{2}D^2f|_a (h,h)\). Clearly \(\phi_h\) is continuous on \([0,1]\) and differentiable on \((0,1)\) with \[\phi_h'(t) = Df|_{a+th}(h) - Df|_a(h) -tD^2 f|_a(h,h)\] Now \( \frac{f(a+h) - f(a) - Df|_a (h) - \frac{1}{2}D^2 f|_a(h,h)}{||h||^2} = \frac{|\phi_h(1) - \phi_h(0)}{||h||^2} = \frac{|\phi_h'(t_h)}{||h||^2}\) for some \(t_h \in (0,1) \)
This is equal to \[\frac{|Df|_{a+t_h} - Df|_a(h) - t_h D^2 f|_a(h,h)|}{||h||^2} = \frac{|D^2 f|_a (t_h h,h) + \alpha_{t_h h}(h) - t_hD^2 f|_a (h,h)}{||h||^2} = \frac{|\alpha_{t_h h}(h)|}{||h||^2}\] where for reach \(u, \alpha_u \in L(\mathbb{R}^n, \mathbb{R})\) with \(\alpha_u = o(u)\) as \(u \to 0\). So \[\frac{ |\alpha_{t_h h}(h)|}{||h||^2} \leq \frac{||\alpha_{t_h h} || || h||}{||h||^2} = \frac{||\alpha_{t_hh}||}{||h||} \leq \frac{||\alpha_{t_hh}||}{||t_h h||} \to 0\] as \(h \to 0\) since \(\alpha_u = o(u)\) and \(t_h h \to 0\) as \(h \to 0\).

Q: Definition. Let \(f : \mathbb{R}^n \to \mathbb{R}, a \in \mathbb{R}^n\). We say a is a local maximum (respectively minimum) of f if [...]
A: there is a neighbourhood N of a s.t. \[\forall x \in N \; f(x) \leq f(a) (\text{respectively} f(x) \geq f(a))\]

Q: Corollary. Let \(f: \mathbb{R}^n \to \mathbb{R}\) be twice continuously differentiable at a with \(Df|_a\) the zero map and \(D^2 f|_a\) positive-definite (respectively negative-definite). Then a is a local minimum (respectively maximum) for f.
A: WLOG \(D^2f|_a\) positive-definite. Then by the second-order Taylor theorem we have \[f(a+h) = f(a) + \frac{1}{2} D^2 f|_a (h,h) + o(||h||^2)\] As \(D^2 f|_a\) is positive-definite, there exists an orthonormal basis of \(\mathbb{R}^n\) with respect to which \(D^2 f|_a\) has diagonal matrix with non-zero entries \(0 < \mu_1 \leq \mu_2 \leq \dots \leq \mu_n\). Hence \[D^2 f|_a (h,h) \geq \mu_1 ||h||^2\]. Take h sufficiently small and non-zero such that the \(o(||h||^2)\) term is at most \(\frac{1}{4}\mu_1 ||h||^2\) in modulus. Then \[f(a+h) \geq f(a) + \frac{1}{2}\mu_1 ||h||^2 - \frac{1}{4}\mu_1 ||h||^2 \geq f(a)\] SO a is a local minimum as claimed.

Q: Theorem. Inverse function Theorem. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\), \(a \in \mathbb{R}^n\) and \(b = f(a)\). Suppose there is some open neighbourhood A of a s.t. f is continuously differentiable at every \(x \in A\) and \(D f|_a\) is non-singular. Then there are open neighbourhoods U of a and V of b such that f is a homeomorphism of U onto V, and if \(g: V \to U\) is the inverse function then g is continuously differentiable at all \(y \in V\) with \(D g|_y = (D f|_a)^{-1}\) where \(f(x) = y\).
A: Let \(\alpha = D f|_a\) and write \[f(a+h) = f(a) + \alpha(h) + \epsilon(h)||h||\] where \(\epsilon(h) \to 0\) as \(h \to 0\).
Let \(\delta, \nu > 0\) be small, to be specified later, but such that if \(x \in \overline{B_{\delta}(a)}\) then f is continuously differentiable at x with \(Df|_x \neq 0\).
Define \(\psi : \overline{B_{\delta}(a)} \to \mathbb{R}^n\) by \(\psi = f - \alpha\). Then \(\psi\) is differentiable at every \(x \in \overline{B_{\delta}(a)}\) with \(D \psi |_x = D f |_x - \alpha \to 0\) as \(x \to a\).
For each \(y \in B_{\nu}(b)\), define \(T_y : \overline{B_{\delta}(a)} \to \mathbb{R}^n\) by \[T_y(x) = x + \alpha^{-1}(y - f(x))\] For any \(x \in \overline{B_{\delta}(a)}\) we have \[ ||T_y(x) - a|| = ||x + \alpha^{-1}(y - f(x)) - a|| = ||\alpha^{-1}(\alpha x + y - f(x) - \alpha a) || \leq ||\alpha^{-1}|| || \alpha x + y - f(x) - \alpha a || = || \alpha^{-1} || || y - f(a) - \epsilon(x-a) ||x - a|| || \leq ||\alpha^{-1}|| (||y-b|| + ||\epsilon(x-a)|| ||x-a||) \leq ||\alpha^{-1}|| (\nu + ||\epsilon(x-a)|| \delta)\]
Also for any \(w,x \in \overline{B_{\delta}(a)}\), we have \[||T_y(w) - T_y(x)|| = || w - \alpha^{-1}(f(w)) - x + \alpha^{-1}(f(x))|| \leq ||\alpha^{-1}|| || \alpha w - f(w) - \alpha x + f(x)|| = || \alpha^{-1} || || \psi(x) - \psi(w) || \leq ||\alpha^{-1}|| || x- w || \sup_{z \in \overline{B_{\delta}(a)}} || D \psi|_z ||\]
Pick \(\delta\) sufficiently small such that \[\forall z \in \overline{B_{\delta}(a)} || D \psi |_z || \leq \frac{1}{2||\alpha^{-1}||}\] and \(x \in \overline{B_{\delta}(a)} \implies ||\epsilon(x-a)|| < \frac{1}{2 ||\alpha^{-1}||}\) and set \(y = \frac{\delta}{2 ||\alpha^{-1}||}\). Then these tell us that \(T_y\) is a contraction of \(\overline{B_{\delta}(a)}\) with \(T_y(\overline{B_{\delta}(a)}) \subset B_{\delta}(a)\) and \(\forall w,x \in \overline{B_{\delta}(a)} \; \; || T_y w - T_y x || \leq \frac{1}{2} || w - x||\).
But \(\overline{B_{\delta}(a)}\) is complete and non-empty so by CMT \(T_y\) has a unique fixed point x and indeed \(x \in B_{\delta}(a)\). Note \(T_y(x) = x \iff f(x) = y\). Let \(V = B_{\nu}(b)\) and let \(U = f^{-1}(B_{\nu}(b)) \cap B_{\delta}(a)\). Then f is continuous bijection of U onto V with U,V open neighbourhoods of a,b respectively.
Now let \(X = C(V, \overline{B_{\delta}(a)}) = \{g : V \to \overline{B_{\delta}(a)} | g \text{ continuous}\}\) with uniform metric \(d(g,h) = \sup_{v \in V} || g(v) - h(v)||\). Then X is non-empty and complete : indeed, suppose \((g_n)\) is a Cauchy sequence in \((X,d)\). Then \((g_n)\) is uniformly Cauchy so uniformly convergent to some \(g: V \to \mathbb{R}^n\). But then also \(g_n \to g\) point wise and \(\overline{B_{\delta}(a)}\) is closed so in fact \(g : V \to \overline{B_{\delta}(a)}\). Finally, a uniform limit of continuous functions is continuous so \(g \in X\), and \(d(g_n , g) \to 0\) as \(n \to \infty\).
For \(g \in X\), define \(T_g : V \to \mathbb{R}^n\) by \(T_g(y) = T_y (g(y))\). From earlier each \(T_y : V \to \overline{B_{\delta}(a)}\) so in fact \(T_g : V \to \overline{B_{\delta}(a)}\). Also \(T_g(z) = T_z(g(z)) = g(z) + \alpha^{-1}(z - f(g(z))) \to g(y) + \alpha^{-1}(y - f(g(y))) = T_y(g(y)) = T_g(y)\) as \(z \to y\), since \(f,g, \alpha^{-1}\) continuous. So \(T_g \in X\). So \(T : X \to X\). Moreover, if \(g,h \in X\) and \(y \in V\) then \[||T g(y) - T h(y)|| = ||T_y(g(y)) - T_y(h(y)) || \leq \frac{1}{2} ||g(y) - h(y)|| \leq \frac{1}{2}d(g,h)\] So \(d(Tg,Th) \leq \frac{1}{2}d(g,h)\). So T is a contraction of X so has a unique fixed point \(g \in X\) by CMT. Now g is a continuous function. And if \(y \in V\), \[g(y) = Tg(y) = T_y(g(y))\] so \(g(y)\) is the unique fixed point of \(T_y\), i.e. the unique \(x \in B_{\delta}(a)\) with \(f(x) = y \). Hence \(g = (f|_U)^{-1}\) so \((f |_U)^{-1}\) is continuous. Thus f is a homeomorphism of U onto V.