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) \ $