Math4201 Topology I (Lecture 15)

Let $U_n=B_{\frac{1}{n}}(x)$ be a sequence of open neighborhoods of $x$.

Since $x\in \overline{A}$ and $U_n$ is an open neighborhood of $x$, then $U_n\cap A\neq \emptyset$. Take $x_n\in U_n\cap A$, we claim that $\{x_n\}_{n=1}^\infty\subseteq A$ that $x_n\to x$.

Take an open neighborhood $U$ of $x$. Then by the definition of metric topology, there is $r>0$ such that $B_r(x)\subseteq U$.

let $N$ be such that $\frac{1}{N}<r$. Then for an $n\geq N$, we have $\frac{1}{n}\leq \frac{1}{N}<r$. This implies that $B_{\frac{1}{n}}(x)\subseteq B_r(x)\subseteq U$.

So $x_n\in B_{\frac{1}{n}}(x)\subseteq U$ for all $n\geq N$.

Otherwise, the last proposition holds for $\mathbb{R}^\omega$ with the box topology.

Last time we showed that this is not the case.

Find an example of a function $f:X\to Y$ which is not continuous but for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$.

Consider $X=\mathbb{R}$ with complement finite topology and $Y=\mathbb{R}$ with the standard topology.

Take identity function $f(x)=x$.

This function is not continuous by trivially taking $(0,1)\subseteq \mathbb{R}$ and the complement of $(0,1)$ is not a finite set, so the function is not continuous.

However, for every convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ trivially.

Part 1:

Let $f:X\to Y$ be a continuous map and

$$\{x_n\}_{n=1}^\infty\subseteq X$$

converges to $x$.

Want to show that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.

i.e. for any open neighborhood $U$ of $f(x)$, we want to show $f(x_n)$ is eventually in $U$. Take $f^{-1}(U)$. This is an open neighborhood of $x$ since $x_n\to x$.

There is $N$ such that $\forall n\geq N$, we have $x_n\in f^{-1}(U)$, $f(x_n)\in U$.

This implies that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.

Part 2:

Let $f:X\to Y$ be a map between two topological spaces with $X$ being metric such that for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$ in $X$, we have $f(x_n)\to f(x)$ in $Y$.

Want to show that $f$ is continuous.

Recall that it suffice to show that for any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.

Take $y\in f(\overline{A})$. Then $y=f(x)$ with $x\in \overline{A}$. By the previous proposition, there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ (since $X$ is a metric space) such that $x_n\to x$.

By our assumption,

$$\{f(x_n)\}_{n=1}^\infty\to f(x) \tag{*}$$

Note that $\{f(x_n)\}_{n=1}^\infty$ is a sequence in $f(A)$ and ($*$) implies that $y=f(x)\in f(A)$ (by the first part of the proposition). This gives us the claim.