User:J D Bowen/Math710 HW2

1) We aim to show that a function $$f:E\subset(X,\sigma)\to(Y,\rho) \ $$ is continuous if and only if $$\lim_{n\to\infty} f(x_n)=f(x) \ $$ whenever $$\left\{{x_n}\right\} \subset E \ $$ and $$x_n \to x \ $$.

($$\Rightarrow$$)

Assume $$f \ $$ is continuous.

Then, given any $$\epsilon>0, \ \exists \delta>0: \ $$
 * $$\sigma(x,y)<\delta \implies \rho(f(x),f(y))<\epsilon \ $$.


 * $$\exists N \ \text{s.t.} \ n>N\implies \sigma(x_n,x)<\epsilon \ $$

Therefore, $$\forall \epsilon, \exists N \ \text{s.t.} \ n>N\implies \rho(f(x_n),f(x))<\epsilon \ $$.

So $$f(x_n)\to f(x) \ $$.

($$\Leftarrow$$)

Let \lim_{n\to\infty} f(x_n) = f(x) for $$\left\{{x_n}\right\} \subset E \ $$ and $$x_n\to x \ $$.

Suppose $$f \ $$ is not continuous. Then $$\exists \epsilon>0 \ $$ such that $$\forall \delta>0, \sigma(x,x_n)<\delta \ $$ but $$\rho(f(x_n),f(x))\geq \epsilon \ $$.

2) Let $$E\subset X \ $$ be a compact subset of a metric space and let $$f:E\to\mathbb{R} \ $$ be continuous. We aim to show $$f \ $$ is uniformly continuous.

work
3) Let $$(X,d) \ $$ be a metric space and $$G\subseteq E \subseteq X \ $$. We say $$G \ $$ is relatively open in $$E \ $$ if $$G=O\cap E \ $$ for some open subset $$O \ $$ of $$X \ $$.  We aim to show that $$G \ $$ is relatively open in $$E \ $$ if and only if $$\forall x \in G, \exists \delta>0 :B_\delta(x)\cap E \subseteq G \ $$.

Define a function $$\text{int}:2^X\to 2^X \ $$ so that $$\text{int}(A) = \left\{{x\in A : \exists \epsilon \ \text{with} \ B_\epsilon(x)\subseteq A }\right\} \ $$.

Since $$G\subseteq E \ $$,

stop
4) Let $$(X, d), (Y, \rho) \ $$ be metric spaces. Suppose $$E \subseteq X, \ f : E \to Y \ $$. Prove that f is continuous if $$f^{-1} (U)$$ is relatively open in E whenever U is open in Y.

working on now
6) Let $$(X, d) \ $$ be a metric space with $$\varnothing \neq A \subseteq X \ $$. Define $$f : X \to R \ $$ by

$$f(x) = \text{inf}_{a\in A} d(x,a) \ $$.

(a) Suppose $$0=f(x)=\text{inf}_{a\in A} d(x,a) \ $$. This is equivalent to stating that $$\forall \epsilon>0, \ \exists a\in A \ $$ such that $$d(x,a)<\epsilon \ $$. If we define $$\epsilon_n :\mathbb{N}\to \mathbb{R}^+ \ $$ as $$\epsilon_n=n^{-1} \ $$, then we can use this (along with the axiom of choice) to pick a series of elements $$a_n\in A \ $$ such that $$d(a_n,x)<\epsilon_n \ $$. Of course, this creates a sequence which converges to $$x \ $$. Hence, the zero set $$\mathbb{V}(f)=\overline{A} \ $$.

(b) We aim to show that $$f:X\to\mathbb{R} \ $$ is a continuous function. If we can show that the image of a sequence converges to the image of the limit of the sequence, then we can apply exercise one. Let $$\epsilon>0 \ $$, and let $$a\in \overline{A} : d(x,a)=f(x) \ $$. Then we have $$\forall x_n \in B_{1/n}(x)\subset X, f(x_n)=d(x_n,a)\leq d(x,x_n)+d(x,a)< 1/n+d(x,a)=1/n+f(x) \ $$ and $$ f(x)=d(x,a)\leq d(x,x_n)+d(x_n,a)< 1/n+d(x_n,a)=1/n+f(x_n) \ $$. Hence $$\forall y \in B_{1/n}(x)\subset X, \ |f(x_n)-f(x)|<\epsilon \ $$. Hence, $$f \ $$ is continuous.

the rest
7) Let $$(X, d) \ $$ be an incomplete metric space. Let $$\hat{X} \ $$ denote the set of all Cauchy sequences in $$X \ $$. Define a relation $$\sim \ $$ on $$\hat{X} \ $$ by $$\left\{{x_n}\right\} \sim \left\{{y_n}\right\} \iff \lim_{n\to\infty} d(x_n,y_n)=0 \ $$.

(a) Show that this an equivalence relation.

We obviously have reflexivity, since $$d(a,a)=0 \ $$ by the definition of metric. We also obviously have symmetry, because all metrics satisfy $$d(a,b)=d(b,a) \ $$.

Finally, suppose $$\left\{{x_n}\right\} \sim \left\{{y_n}\right\} \ $$ and $$\left\{{y_n}\right\} \sim \left\{{z_n }\right\} \ $$. This means we have $$\lim_{n\to\infty} d(x_n,y_n)=0 \ $$ and $$\lim_{n\to\infty} d(y_n,z_n)=0 \ $$. But then by the triangle inequality, $$\lim_{n\to\infty} d(x_n,z_n) \leq \lim_{n\to\infty} d(x_n,y_n)+d(y_n,z_n)=\lim_{n\to\infty} d(x_n,y_n)+\lim_{n\to\infty} d(y_n,z_n)=0+0=0 \ $$. So $$\left\{{x_n}\right\} \sim \left\{{z_n}\right\} \ $$.

now
(b) Let $$X^* \ $$ denote the equivalence classes on $$\hat{X} \ $$ defined by $$\sim \ $$. For $$[\left\{{x_n}\right\}], [\left\{{y_n}\right\}] \in X^* \ $$, define

$$d^* ([\left\{{x_n}\right\}], [\left\{{y_n}\right\}]) = \lim_{n\to\infty} d(x_n,y_n) \ $$.

Show $$(X^*, d^*) \ $$ is a complete metric space.

Let Greek letters denote equivalence classes, $$\alpha\in X^* \ $$ and Latin letters, point in $$x \in X \ $$. We aim to show that for every Cauchy sequence $$\left\{{\alpha_1, \alpha_2, \dots }\right\} \subset X^* \ $$, there exists a $$\alpha\in X^* \ $$ such that $$\alpha_n \to \alpha \ $$.

Since each alpha is an equivalence class of Cauchy sequences, there is a Cauchy sequence in each. Pick one and call it $$\left\{{x_1^n, x_2^n, \dots }\right\} \in \alpha_n \ $$.

Then we can form the sequence $$\left\{{x_1^1, x_2^2, x_3^3, \dots }\right\} \subset X \ $$.


 * Lemma: This sequence is Cauchy.
 * Proof:

Since this sequence is Cauchy, it belongs to an equivalence class $$\phi \in X^* \ $$. Observe that

$$\lim_{n\to\infty} d^*(\alpha_n, \phi) = \lim_{n\to\infty} \lim_{m\to\infty} d(x^n_m, x^m_m) = \lim_{m\to\infty} \lim_{n\to\infty} d(x^n_m, x^m_m) \ $$.

Since the alpha sequence is Cauchy, and each representative sequence from an alpha is Cauchy, given any $$\epsilon > 0, \ \exists M \ $$ such that $$\forall m,n,p,q>M \ $$, we have $$d(x^m_n, x^p_q )< \epsilon \ $$.

So, $$\alpha_n\to [\left\{{x^n_n}\right\}] \ $$.

later
(c) Define the map $$j:X\to X^* \ $$ by $$j(x)=[\left\{{x_n}\right\}] \ $$ where $$x_n=x \ \forall n \ $$. Show that $$d^*(j(x),j(y)) = d(x,y) \ $$ and $$\overline{j(X)}=X^* \ $$.

$$d^*(j(x),j(y)) = d^* ([\left\{{x_n}\right\}], [\left\{{y_n}\right\}]) = \lim_{n\to\infty} d(x_n,y_n)=d(x,y) \ $$