# Math4201 Topology I (Lecture 34)

## Countability axioms

### Second countability axiom

<details>
<summary>Example of spaces that is first countable but not second countable</summary>

Let $\mathbb{R}_l$ be $\mathbb{R}$ with the lower limit topology generated by the basis:

$$
\mathcal{B}=\{[a,b)\mid a,b\in \mathbb{R},a<b\}
$$

And $\mathbb{R}_l$ is first countable since for each $x\in \mathbb{R}$, there is a countable collection $\{[x,\frac{1}{n}+x)\}_{n\in \mathbb{N}}$ of open intervals in $\mathbb{R}_l$ such that any open interval $U$ of $x$ contains one of $[x,\frac{1}{n}+x)\}$

However, $\mathbb{R}_l$ is not second contable:

If $\mathbb{R}_l$ is second countable, then for any real number $x$, there is an element $U$ of $\mathcal{B}$ contains $x$ and is contained in $[x,x+1)$.

Any such open sets is of the form $[x,x+\epsilon)\cap A$ with $\epsilon>0$ and any element of $A$ being larger than $\min(U_x)=x$.

In summary, for any $x\in \mathbb{R}$, there is an element $U_x\in \mathcal{B}$ with $(U_x)=x$. In particular, if $x\neq y$, then $U_x\neq U_y$. SO there is an injective map $f:\mathbb{R}\rightarrow \mathcal{B}$ sending $x$ to $U_x$. This implies that $\mathbb{B}$ is uncountable.

</details>

#### Proposition of second countable spaces

Let $X$ be a second countable topological space. Then the following holds:

1. Any discrete subspace $Y$ of $X$ is countable
2. There exists a countable subset of $X$ that is dense in $X$ (_also called separable spaces_)
3. Every open covering of $X$ has **countable** subcover (That is if $X=\bigcup_{\alpha\in I} U_\alpha$, then there exists a **countable** subcover $\{U_{\alpha_1}, ..., U_{\alpha_\infty}\}$ of $X$) (_also called Lindelof spaces_)

<details>
<summary>Proof</summary>

First we prove that any discrete subspace $Y$ of $X$ is countable.

Let $Y$ be a discrete subspace of $X$. In particular, for any $y\in Y$ we can find an element $B_y$ of the countable basis $\mathcal{B}$ for $Y$ such that $B_y\cap Y=\{y\}$.

In particular, if $y\neq y'$, then $B_y\neq B_{y'}$. Because $\{y\}=B_y\cap Y\neq B_{y'}\cap Y=\{y'\}$.

This shows that $\{B_y\}_{y\in Y}\subseteq B$ has the same number of elements as $Y$.

So $Y$ has to be countable.

---

Next we prove that there exists a countable subset of $X$ that is dense in $X$.

For each basis element $B\in \mathcal{B}$, we can pick an element $x\in B$ and let $A$ be the union of all such $x$.

We claim that $A$ is dense.

To show that $A$ is dense, let $U$ be a non-empty open subset of $X$.

Take an element $x\in U$. Note that by definition of basis, there is some element $B\in \mathcal{B}$ such that $x\in B$. So $x\in B\cap U$. $U\cap B\neq \emptyset$, so $A\cap U\neq \emptyset$.

Since $A\cap U\neq \emptyset$ this shows that $A$ is dense.

---

Then we prove that every open covering of $X$ has countable subcover.

Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $X$. Let $\mathcal{B}$ be a countable basis for $X$.

For any basis element $B$ of $X$. If $B$ is in $U_\alpha$ for some $\alpha\in I$, then pick $U_\alpha$ as an element of our subcover.

This way we get countably many open sets $U_\alpha$'s because $\mathcal{B}$ is countable.

We also claim that the chosen $U_\alpha$'s given an open covering of $X$.

For any $x\in X$, there is $U_\alpha$ (possibly not one of the chosen ones) such that $x\in U_\alpha$. There is $B_x\in \mathcal{B}$ such that $x\in B_x\subseteq U_\alpha$.

In particular, there is a chosen $U_\alpha$ such that $B_x\subseteq U_\alpha$. This implies that there is a chosen $U_\alpha$, containing $x$.

</details>

### Separation Axioms

> Our goal is to find conditions that if some space is second countable, and xxx, then it's metrizable.

#### Kolmogorov classification

Let $X$ be a topological space:

- $X$ is $T_0$ if for any pair of points $x,y\in X$, $x\neq y$, there is an open set $U$ containing $x$ but not $y$. (equivalent to say that any singleton set is closed)
- $X$ is Hausdorff if for any pair of distinct $x,y\in X$, there are **disjoint** open sets $U$ and $V$ such that $x\in U$ and $y\in V$.

#### Definition of regular spaces

A $T_0$ space is regular if for any $x\in X$ and any close set $A\subseteq X$ such that $x\notin A$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $A\subseteq V$.

#### Definition of normal spaes

A $T_0$ space is normal if for any disjoint closed sets, $A,B\subseteq X$, there are **disjoint open sets** $U,V$ such that $A\subseteq U$ and $B\subseteq V$.

<details>
<summary></summary>

Let $\mathbb{R}_K$ be the topology on $\mathbb{R}$ generated by the basis:

$$
\mathcal{B}=\{(a,b)\mid a,b\in \mathbb{R},a<b\}\cup \{(a,b)-K\mid a,b\in \mathbb{R},a<b\}
$$

where $K=\coloneqq \{\frac{1}{n}\mid n\in \mathbb{N}\}$.

**This is finer than the standard topology** on $\mathbb{R}$.

Since $(-1,1)-K$ is not open in $\mathbb{R}$, but it is open in $\mathbb{R}_K$.

$\mathbb{R}_K$ is Hausdorff, since for any $x\neq y$, there is an open set $(x-\epsilon,x+\epsilon)$ and $(y-\epsilon,y+\epsilon)$ such that $(x-\epsilon,x+\epsilon)\cap (y-\epsilon,y+\epsilon)=\emptyset$.

However, this space is not regular.

consider $x=0$ and $A=K$ is a closed set.

Any open neighborhood $U$ of $x$ and $V$ of $A$ are not disjoint.

Otherwise, if there exists such $U$ and $V$, we can assume $U$ is a basis element. If $U=(a,b)$ with $0\in U$ then $U\cap K\neq \emptyset$.

Which is a contradiction.

So $U=(a,b)-K$. Suppose $\frac{1}{n}\in (a,b)$.

Then $V$ contains an open interval of the form $(\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)$. But $(a,b)-K\cap (\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)\neq \emptyset$.

Which is a contradiction.

</details>

Let $X$ be a regular space. Take $x\in X$ and an open neighborhood $U$ of $x$. So $A\coloneqq X-U$ is a closed set disjoint from $U$.

By regularity assumption, there is an open neighborhood $W_1$ of $x$ and $W_2$ of $A$ that are disjoint.

In particular, $X-W_2$ is a closed set contained in $X-A$.

In particular, $X-W_2$ is a closed set contained in $X-A=U$ which also contains $W_1$. This implies that the closure of $W_1$ is contained in $U$.

#### Lemma of regular spaces

If $X$ is a regular space, and $x\in X$, and $U$ is an open neighborhood of $x$, then there is an open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.

Continue next lecture.