# Math4201 Lecture 6

## Product topology

### Define topological spaces on cartesian product of two topological spaces

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.

$$
X\times Y=\{(x,y)|x\in X,y\in Y\}
$$

Goal: Define a topology on $X\times Y$.

One way is to take $\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}$ is a basis for the topology on $X\times Y$.

> There are two ways to define the topology on $\mathbb{R}^2$: By rectangles ($\mathcal{B}_{rect}=\{(a,b)\times (c,d)|a,b,c,d\in \mathbb{R},a<b,c<d\}$) or by open disks ($\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$). (check bonus video)

<details>
<summary>Proof</summary>

Take $U=X,V=Y$, then $(x,y)\in (U\times V)=X\times Y$.

So the first property of basis is satisfied.

Check the second property of basis:

Let $B_1=U_1\times V_1,B_2=U_2\times V_2$ be two basis elements, and $(x,y)\in B_1\cap B_2$.

$$
B_1\cap B_2=(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)
$$

Since $U_1\cap U_2\in \mathcal{T}_X$ and $V_1\cap V_2\in \mathcal{T}_Y$, we have $(U_1\cap U_2)\times (V_1\cap V_2)\in \mathcal{B}_{X\times Y}$.

Take $B_3=(U_1\cap U_2)\times (V_1\cap V_2)$, then $(x,y)\in B_3=B_1\cap B_2$.

</details>

> [!CAUTION]
>
> $\mathcal{B}_{X\times Y}$ is not a topology on $X\times Y$.
>
> $\mathcal{B}_{X\times Y}$ is not closed with respect to arbitrary unions.

#### Definition of product topology

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.

$$
\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}
$$

The product topology or $X\times Y$ is the topology generated by $\mathcal{B}_{X\times Y}$.

Let $\mathcal{B}_X$ be a basis for $\mathcal{T}_X$ and $\mathcal{B}_Y$ be a basis for $\mathcal{T}_Y$.

If we define

$$
\mathcal{B}'_{X\times Y}=\{(U\times V)|U\in \mathcal{B}_X,V\in \mathcal{B}_Y\}
$$

#### Proposition

$\mathcal{B}'_{X\times Y}$ is a basis for the product topology on $X\times Y$.

> [!NOTE]
>
> This basis is smaller than $\mathcal{B}_{X\times Y}$.
>
> Consider $X=Y=\mathbb{R}$ and $U=(a,b)\cap (c,d)$ and $V=(e,f)\cap (g,h)$. (Assume $a<b,c<d,e<f,g<h$) The union of $U$ and $V$ is four rectangles, which is not in $\mathcal{B}'_{X\times Y}$. but it is in $\mathcal{B}_{X\times Y}$.

<details>
<summary>Proof</summary>

Using the [lemma](https://notenextra.trance-0.com/Math4201/Math4201_L4#theorem-of-basis-of-topology) from Friday. it suffices to show that:

Let $W\in X\times Y$ and $(x,y)\in W$, we need to show that there are $B\in \mathcal{B}_X, B'\in \mathcal{B}_Y$ such that $(x,y)\in (B\times B')\subseteq W$.

Since $W$ is open with respect to the topology generated by $\mathcal{B}_{X\times Y}$, in particular, there is $U\times V$ such that $(x,y)\in U\times V\subseteq W$. And $x\in U$ and $y\in V$.

Since $U\in \mathcal{B}_X$ and $V\in \mathcal{B}_Y$, by property of basis $\mathcal{B}_X$ and $\mathcal{B}_Y$, $\forall x\in U$, $\exists B\in \mathcal{B}_X$ such that $x\in B\subseteq U$ and $\forall y\in V$, $\exists B'\in \mathcal{B}_Y$ such that $y\in B'\subseteq V$.

So $(x,y)\in (B\times B')\subseteq U\times V\subseteq W$.

</details>

#### Definition of standard topology on $\mathbb{R}^n$

The standard topology on $\mathbb{R}^n$ is defined as the product topology on $\mathbb{R}^{n-1}\times \mathbb{R}$.

## Subspace topology

Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$.

Want to define a topology on $Y$.

$$
\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}
$$

We claim that $\mathcal{T}_Y$ is a topology on $Y$, called as the subspace topology on $Y$.

<details>
<summary>Proof</summary>

First, $\emptyset \cap Y=\emptyset \in \mathcal{T}_Y$ and $Y=X\cap Y\in \mathcal{T}_Y$.

Second, let $\{U_\alpha\cap Y\}_{\alpha \in I}$ be collection of open sets in $\mathcal{T}_Y$. Note that $U_\alpha\in \mathcal{T}$ for all $\alpha \in I$.

So, $\bigcup_{\alpha \in I} U_\alpha\cap Y=\left(\bigcup_{\alpha \in I} U_\alpha\right)\cap Y\in \mathcal{T}_Y$ because $\bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}$.

Third, let $\{U_i\cap Y\}_{i=1}^n$ be a finite collection of open sets in $\mathcal{T}_Y$. Note that $U_i\in \mathcal{T}$ for all $i=1,2,\ldots,n$.
So, $\bigcap_{i=1}^n U_i\cap Y=\left(\bigcap_{i=1}^n U_i\right)\cap Y\in \mathcal{T}_Y$ because $\bigcap_{i=1}^n U_i\in \mathcal{T}$.

</details>

### Generate basis for subspace topology

Let $\mathcal{B}$ is a basis for $(X,\mathcal{T})$. We'd like to use $\mathcal{B}$ to define a basis for $(Y,\mathcal{T}_Y)$.

$$
\mathcal{B}_Y=\{B\cap Y|B\in \mathcal{B}\}
$$

#### Proposition for basis of subspace topology

$\mathcal{B}_Y$ is a basis for a topology on $Y$ that generates the subspace topology on $Y$.

Proof as exercise. (same as the proof for the basis of product topology)

<details>
<summary>Example: not every open set in subspace topology is open in the original space</summary>

Let $X=\mathbb{R}$ with standard topology and $Y=[0,1]\cup [2,3]$. equipped with subspace topology generated by the standard basis for $\mathbb{R}$.

so $[0,1]=(-1,\frac{3}{2})\cap Y$ In particular, $[0,1]$ is open set in $Y$, but not an open set in $\mathbb{R}$.

</details>

#### Lemma of open set in subspace topology

Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$ is open. $Z\subseteq Y$ is open subset with respect to the subspace topology on $Y$. Then $Z$ is open in $X$.