Math4201 Lecture 16 (Topology I)

If $(Y,d)$ is a bounded metric space, let $M$ be a positive constant, then $\overline{d}=\min\{M,d\}$ is a bounded metric space.

In fact, the metric topology by $d$ and $\overline{d}$ are the same. (proved in homeworks)

Proof is similar to showing that the square metric is a metric on $\mathbb{R}^n$.

$\rho(f,g)=0\implies \sup_{x\in X}(d(f(x),g(x)))=0$

Since $d(f(x),g(x))\geq 0$, this implies that $d(f(x),g(x))=0$ for all $x\in X$.

The triangle inequality of being metric for $\rho$ follows from the similar properties for $d$.

We need to show that $\overline{Z}=Z$.

Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: Let $f_n:X\to Y\in Z$ be a sequence of continuous maps,

Which is to prove the uniform convergence,

$$f_n \to f \in \operatorname{Map}(X,Y)$$

Then we want to show that $f$ is also continuous.

It is to show that for any open subspace $V$ of $Y$, $f^{-1}(V)$ is open in $X$.

Take $x_0\in f^{-1}(V)$, we’d like to show that there is an open neighborhood $U$ of $x_0$ such that $U\subseteq f^{-1}(V)$.

Since $x_0\in f^{-1}(V)$, then $f(x_0)\in V$. By metric definition, there is $r>0$ such that $B_r(f(x_0))\subseteq V$.

Take $N$ to be large enough such $\rho(f_N(x), f(x)) < \frac{r}{3}$

So $\forall x\in X$, $d(f(x),f_N(x))<\frac{r}{3}$

Since $f_N$ is continuous, $f_N^{-1}(B_{r/3}(f(x_0)))$ is an open set $U\subseteq X$ containing $x_0$.

Take $x\in U$, $d(f(x),f(x_0))<d(f(x),f_N(x_0))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))$ using triangle inequality.

Note that,

$d(f(x),f_N(x))<\frac{r}{3}$ (using $N$ large enough),

$d(f_N(x),f_N(x_0))<\frac{r}{3}$ (using $x\in U$, then $f_N(x)\in B_{r/3}(f_N(x_0))$, so $d(f_N(x),f_N(x_0))<\frac{r}{3}$),

$d(f_N(x_0),f(x_0))<\frac{r}{3}$ (using $N$ large enough),

So $d(f(x),f(x_0))<\frac{r}{3}+\frac{r}{3}+\frac{r}{3}=r$.

So $f(x)\in B_r(f(x_0))\implies x\in f^{-1}(B_r(f(x_0)))\implies x\in f^{-1}(V)\implies U\subseteq f^{-1}(V)$.

So $f^{-1}(V)$ is open in $X$.