# Math4201 Topology I (Lecture 14)

## Metric topology

### Product topology and metric topology

If $X$ and $Y$ are metrizable spaces, then the product space $X\times Y$ is metrizable.

If $X$ is metrizable, then the subspace $A\subset X$ equipped with subspace topology is metrizable.

<details>
<summary>Proof</summary>

Let $d$ be a metric on $X$. Then define $d'$ be the restriction of $d$ to $A$:

$$
d':A\times A\to \mathbb{R}+
$$

$$
d'(x,y)=d(x,y)
$$

$x,y\in A\subseteq X$

$d'$ is a metric on $A$. Since the metric topology on $A$ associated to $d'$ is the same as the subspace topology.

Note that for any $x\in A$ and $r>0$

$$
B_r^{d}(x)\cap A=B_r^{d'}(x)\tag{*}
$$

A basis for metric topology on $A$ is given by:

$$
\mathcal{B}=\{B_r^{d'}(x)|x\in A,r>0,r\in \mathbb{R}\}
$$

A basis for the subspace topology on $A$ is given by:

$$
\mathcal{B}'=\{B_r^{d}(x)\cap A|x\in A,r>0,r\in \mathbb{R}\}
$$

Since (*) holds, $\mathcal{B}\subseteq \mathcal{B}'$.

This shows that subspace topology on $A$ is finer than the metric topology on $A$.

We need to show that for any $B_r^{d}(x)$ with $x\in X$ and $y\in B_r^{d}(x)\cap A$, we have $r'>0$ such that $y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A$.

Use triangle inequality, we have $r'=r-d(x,y)>0$ such that $y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A$.

</details>

#### Proposition on sequence and closure

Let $X$ be a topological space and $A\subseteq X$. Then the following holds:

If there is a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to x$, then $x\in \overline{A}$. ($x$ may not be in $A$)

The reverse holds if $X$ is a metric space. That is, if $X$ is a metric space and $x\in \overline{A}$, then there is a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to x$.

<details>
<summary>Example of non-metrizable space</summary>

For the second part of the claim

Let $X=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ with the product topology over infinite product.

$X=\text{Map}(\mathbb{N},\mathbb{R})=\{(x_1,x_2,x_3,\cdots)|x_i\in \mathbb{R},i\in \mathbb{N}\}$.

The box topology on $X$ is the topology generated by:

$$
\mathcal{B}=\{(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots|a_i,b_i\in \mathbb{R},i\in \mathbb{N}\}
$$

It is easy to check that this is a basis.

Take $A=\mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i> 0\}$.

$\overline{A}=\mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i\geq 0\}$.

In particular, $(0,0,0,\cdots)\in \overline{A}$

Take a basis element $B=(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots\in \mathcal{B}$ containing $(0,0,0,\cdots)$. This means that $a_i<0<b_i$ Then $(\frac {b_1}{2},\frac {b_2}{2},\frac {b_3}{2},\cdots)\in B\cap A$.

This shows that $(0,0,0,\cdots)\in \overline{A}$.

We claim that there is no sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to (0,0,0,\cdots)$.

We proceed by contradiction and suppose that there is such a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to (0,0,0,\cdots)$.

Since $x_n\in A$, we have $v_n=(a_1^n,a_2^n,a_3^n,\cdots)$ with $a_i^n>0$ for all $i\in \mathbb{N}$.

Consider the following open set around $(0,0,0,\cdots)$:

$C=(-\frac{a_1^1}{2},\frac{a_1^1}{2})\times (-\frac{a_2^1}{2},\frac{a_2^1}{2})\times (-\frac{a_3^1}{2},\frac{a_3^1}{2})\times \cdots$

We claim that $v_n\notin C$. Otherwise, we should have $-\frac{a_i^j}{2}<a_j^n<\frac{a_j^j}{2}$ for all $i\in \mathbb{N}$.

But this doesn't hold for $j=n$. This shows that $v_n\cancel{\to} (0,0,0,\cdots)$. Which is a contradiction.

---

For the first part of the claim. Let $\{x_n\}_{n=1}^\infty\subset A$ converge to $x$. Then for any open set $U$ of $x$, we have $N$ such that $x_n\in U$ for $n\geq N$.

In particular, $U\cap A$ is non-empty because it has $x_n$ for large enough $n$.

So $x\in \overline{A}$.

</details>