# Math4201 Topology I (Lecture 33)
## Countability axioms
### First countability axiom
For any $x\in X$, there is a countable collection $\{B_n\}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$.
Example: Metric spaces
### Second countability axiom
There exists countable basis for a topology on $X$.
Example: $\mathbb{R}$ and more generally $\mathbb{R}^n$
Consider the following topology on $\mathbb{R}^\omega$:
There exists different topology can be defined on $\mathbb{R}^\omega$
- Product topology
- Box topology
- Union topology
#### Definition of product topology
Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces:
$$
\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\}
$$
The product topology defined on the basis that:
The set of following forms:
$$
\mathcal{B}_{prod}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open and }U_\alpha=X_\alpha\text{ for all except finitely many }\alpha\}
$$
So $\mathbb{R}^\omega$ with product topology is second countable.
Proof that product topology defined above is second countable
A countable basis for $\mathbb{R}^\omega$ is given by the union of following sets:
$$
B_1=\{(a_1,b_1)\times \mathbb{R}\times \mathbb{R}\times \dots |a_1,b_1\in \mathbb{Q}\}\\
B_2=\{(a_1,b_1)\times(a_2,b_2) \mathbb{R}\times \dots |a_2,b_2\in \mathbb{Q}\}\\
B_n=\{(a_1,b_1)\times(a_2,b_2)\times \dots (a_n,b_n)\times \mathbb{R}\times \dots |a_n,b_n\in \mathbb{Q}\}\\
$$
Each $B_i$ is countable and there exists a bijection from $B_n\cong\mathbb{Q}^{2n}$
And the union of countably many countable sets is also countable.
So $B_1\cup B_2\cup \dots$ is countable. This is also a baiss for the product topology on $\mathbb{R}^\omega$
#### Lemma of second countable spaces
If $X_1,X_2,X_3,\dots$ are second countable topological spaces, then the following spaces are also second countable:
1. $X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology)
2. $X_1\times X_2\times X_3\times \dots$ with the product topology
Ideas for proof
For $X_1\times X_2\times X_3\times \dots\times X_n$:
$B_1\times B_2\times B_3\times \dots\times B_n$ is also countable basis for $X_1\times X_2\times X_3\times \dots\times X_n$
Let $\tilde{B}\coloneqq B_1\times B_2\times B_3\times \dots\times B_i\times X_{i+1}\times X_{i+2}\times \dots$ then $\tilde{B}$ is a countable basis for $X_1\times X_2\times X_3\times \dots$
#### Lemma of first countable spaces
If $X_1,X_2,X_3,\dots$ are first countable topological spaces, then the following spaces are also first countable:
1. $X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology)
2. $X_1\times X_2\times X_3\times \dots$ with the product topology
Ideas for proof
Basically the same as before but now you have analyze the basis for each $x\in X_1\times X_2\times X_3\times \dots$
#### Definition of box topology
Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces:
$$
\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\}
$$
The product topology defined on the basis that:
The set of following forms:
$$
\mathcal{B}_{box}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open}\}
$$
#### Definition of uniform topology
The uniform topology on $X$ is the topology induced by the uniform metric on $X$.
$$
\rho(x,y)=\sup_{i\in \mathbb{N}}\overline{d}(x_\alpha,y_\alpha)
$$
To get a finite number for $\rho(x,y)$, we define the bounded metric $\overline{d}$ on $X$ by $\overline{d}(x,y)=\min\{d(x,y),1\}$ where $d$ is the usual metric on $X$.
where $x=(x_1,x_2,x_3,\dots), y=(y_1,y_2,y_3,\dots)\in X$
In particular, the $\mathbb{R}^\omega$ with the uniform topology is first countable because it's a metric space.
However, it's not second countable.
> Recall that $Y\subseteq X$ is a discrete subspace if the subspace topology on $Y$ is the discrete topology, i.e. any point of $y$ is open in $Y$.
Define $A\subseteq \mathbb{R}^\omega$ be defined as follows:
$$
A=\{\underline{x}=(x_1,x_2,x_3,\dots)|x_i\in \{0,1\}\}
$$
Let $\underline{x}$ and $\underline{x}'$ be two distinct elements of $A$.
$$
\rho(\underline{x},\underline{x}')=\sup_{i\in \mathbb{N}}\overline{d}(\underline{x}_\alpha,\underline{x}'_\alpha)=1
$$
(since there exists at least one entry different in $\underline{x}$ and $\underline{x}'$)
In particular, $B_1^\rho(\underline{x})\cap A=\{\underline{x}\}$, so $A$ is a discrete subspace of $\mathbb{R}^\omega$.
This subspace is also uncountable ($A$ can create a surjective map to $(0,1)$ using binary representation) which implies that $\mathbb{R}^\omega$ is not second countable.
#### 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_)
Proof
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.
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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.
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Third part next lecture.