# Math4201 Topology I (Lecture 15)

## Continue on convergence of sequences

### Closure and convergence

#### Proposition in metric space, every convergent sequence converges to a point in the closure

If $X$ is a metric space, $A\subset X$, and $x\in \overline{A}$, then there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$.

<details>
<summary>Proof</summary>

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 $n\geq N$.

</details>

#### Corollary $\mathbb{R}^\omega$ with the box topology is not metrizable

$\mathbb{R}^\omega$ is $\text{Map}(\mathbb{N},\mathbb{R})=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$.

> Note that $\mathbb{R}^\omega$ is Hausdorff. So not all Hausdorff spaces are metrizable.

<details>
<summary>Proof</summary>

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

Last time we showed that this is not the case.
</details>

#### Definition of first countability axiom

A topological space $(X,\mathcal{T})$ satisfies the **first countability axiom** if for any point $x\in X$, there is a sequence of open neighborhoods of $x$, $\{V_n\}_{n=1}^\infty$ such that any open neighborhood $U$ of $x$ contains one of $V_n$.

Note that if we take $U_n=V_1\cap V_2\cap \cdots \cap V_n$, then any open neighborhood $U$ that contains $V_N$, then it also contains $U_n$ for all $n\geq N$.

> As the previous prof, for metric space, it is natural to try $V_n=B_{\frac{1}{n}}(x)$ for some $x\in X$.

#### Proposition on first countability axiom

Rewrite the [Proposition of metric space, every convergent sequence converges to a point in the closure](#proposition-in-metric-space-every-convergent-sequence-converges-to-a-point-in-the-closure)

We can have the following:

If $(X,\mathcal{T})$ satisfies the first countability axiom, then every convergent sequence converges to a point in the closure.

We can easily prove this by takeing the sequence of open neighborhoods $\{V_n\}_{n=1}^\infty$ instead of $U_n=B_{\frac{1}{n}}(x)$.

#### Proposition of continuous functions

Let $f:X\to Y$ be a map between two topological spaces.

1. If $f$ is continuous, then for any convergent sequence $\{x_n\}_{n=1}^\infty$ in $X$ converging to $x$, the sequence $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
2. If $X$ is equipped with the metric topology and for any convergent sequence $\{x_n\}_{n=1}^\infty\to x$ in $X$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ in $Y$, then $f$ is continuous.

<details>
<summary>Exercise</summary>

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

</details>

<details>
<summary>Solution</summary>

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.

</details>

<details>
<summary>Proof</summary>

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.
</details>

> [!NOTE]
>
> The second part of the proposition is also true when $X$ is not a metric space but satisfies the first countability axiom.

#### Equivalent formulation of continuity

If $(X,d)$ and $(Y,d')$ are metric spaces and $f:X\to Y$ is a map, then $f$ is continuous if and only if for any $x_0\in X$ and any $\epsilon > 0$, there exists $\delta > 0$ such that if $\forall n\in X, d(x_n,x_0)<\delta$, then $d'(f(x_n),f(x_0))<\epsilon$.

Proof similar to the $X=Y=\mathbb{R}$ case.