# Math4201 Topology I (Lecture 4)

## Recall from last lecture

Assignment due next Thursday. 10PM

Let $\mathcal{B}$ be a basis for a topology. Then the topology ($\mathcal{T}_{\mathcal{B}}$) **generated** by $\mathcal{B}$ is $\{U\in \mathcal{T}_{\mathcal{B}} \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U\}$.

## New materials

### Topology basis

Given a topology on a set $X$, When is a given collection of subsets of $X$ a basis for a topology?

Suppose $U\in\mathcal{T}$ is an open set in $X$. If an arbitrary set $\mathcal{C}$ is a basis for $\mathcal{T}$, then by the definition of a topology generated by a basis, we should have the following:

$$
\exists C\in \mathcal{C} \text{ such that } x\in C\subseteq U
$$

#### Theorem of basis of topology

> [!CAUTION]
>
> In this course, we use lowercase letters to denote element of a set, and uppercase letters to denote sets. We use $\mathcal{X}$ to denote set of subsets of $X$.

Let $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of subsets of $X$ satisfying the following property:

$$
\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
$$

Then $\mathcal{C}$ is a basis and the topology generated by $\mathcal{C}$ is $\mathcal{T}$.

<details>
<summary>Proof</summary>

We want to show that $\mathcal{C}$ is a basis.

> Recall the definition of a basis:
>
> 1. $\forall x\in X$, there is $B\in \mathcal{B}$ such that $x\in B$
> 2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, there is $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$

First, we want to show that $\mathcal{C}$ satisfies the first property.

Take $x\in X$. Since $X\in \mathcal{T}$, we can apply the given condition ($
\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
$) to get $C\in \mathcal{C}$ such that $x\in C\subseteq X$.

Next, we want to show that $\mathcal{C}$ satisfies the second property.

Let $C_1,C_2\in \mathcal{C}$ and $x\in C_1\cap C_2$. Since $C_1,C_2\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U=C_1\cap C_2\in \mathcal{T}$. 

We can apply the given condition to get $C_3\in \mathcal{C}$ such that $x\in C_3\subseteq U=C_1\cap C_2$.

---

Then we want to show that the topology generated by $\mathcal{C}$ is $\mathcal{T}$.

> Recall the definition of the topology generated by a basis:
>
> To prove this, we need to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$ and $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.
>
> Moreover, from last lecture, we have $U\in \mathcal{T}_{\mathcal{B}}\iff U=\bigcup_{\alpha \in I} B_\alpha$ for some $\{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$.

First, we want to show that $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.

Let $U=\bigcup_{\alpha \in I} C_\alpha$ for some $\{C_\alpha\}_{\alpha \in I}\subseteq \mathcal{C}$. Then since $C_\alpha\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U\in \mathcal{T}$.

Next, we want to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$.

Let $U\in \mathcal{T}$. Then $\forall x\in U$ by the given condition, we have $C\in \mathcal{C}$ such that $x\in C\subseteq U$.

So, $U=\bigcup_{\alpha \in I} C_\alpha\in \mathcal{T}_{\mathcal{C}}$. (using the [same trick last time](https://notenextra.trance-0.com/Math4201/Math4201_L3#lemma))

</details>

Let $\mathcal{T}$ be the topology on $X$. Then $\mathcal{T}$ itself satisfies the basis condition.

#### Definition of subbasis of topology

A subbasis of a topology on a set $X$ is a collection $\mathcal{S}\subseteq \mathcal{T}$ of subsets of $X$ such that their union is $X$.

$$
\mathcal{S}=\{S_{\alpha}\mid S_\alpha\subseteq X\}_{\alpha \in I}\text{  and  }\bigcup_{\alpha \in I} S_\alpha=X
$$

#### Definition of topology generated by a subbasis

If we consider the basis generated by the subbasis $\mathcal{S}$ by the following:

$$
\mathcal{B}=\{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
$$

Then $\mathcal{B}$ is a basis.

<details>
<summary>Proof</summary>

First, $\forall x\in X$, there is $S_\alpha\in \mathcal{S}$ such that $x\in S_\alpha$. In particular, $x\in \mathcal{B}$.

Second, let $B_1,B_2\in \mathcal{B}$. Since $B_1$ is the intersection of a finite number of elements of $\mathcal{S}$, we have $B_1=\bigcap_{i=1}^n S_{i_1}, B_2=\bigcap_{i=1}^n S_{i_2}$ for some $S_{i_1},S_{i_2}\in \mathcal{S}$.

So $B_1\cap B_2$ is the intersection of finitely many elements of $\mathcal{S}$.

So $B_1\cap B_2\in \mathcal{B}$.

</details>

We call the topology generated by $\mathcal{B}$ the topology generated by the subbasis $\mathcal{S}$. Denote it by $\mathcal{T}_{\mathcal{S}}$.

An open set with respect to $\mathcal{T}_{\mathcal{S}}$ is a subset of $X$ such that it can be written as a union of finitely intersections of elements of $\mathcal{S}$.

<details>

<summary>Example (standard topology on real numbers)</summary>

Let $X=\mathbb{R}$. Take $\mathcal{S}=\{(-\infty, a)|a\in \mathbb{R}\}\cup \{(a,+\infty)|a\in \mathbb{R}\}$.

We claim this is a subbasis of the standard topology on $\mathbb{R}$.

The basis $\mathcal{B}$ associated with $\mathcal{S}$ is the collection of all open intervals.

$$
\mathcal{B}=\{(a,b)=(-\infty, b)\cap (a,+\infty)\}
$$

So, $\mathcal{B}=\mathcal{B}_{st}$ (the standard basis).

This topology on $\mathbb{R}$ is the same as the standard topology on $\mathbb{R}$.

</details>

<details>
<summary>Example (finite complement topology)</summary>

Let $X$ be an arbitrary set. Let $\mathcal{S}$ defined as follows:

$$
\mathcal{S}=\{S\subseteq X\mid S=X\setminus \{x\} \text{ for some } x\in X\}
$$

Let $x,y\in X$ and $x\neq y$. Then $S_x=X\setminus \{x\}$ and $S_y=X\setminus \{y\}$ are two elements of $\mathcal{S}$. Since $x\neq y$, we have $S_x\cup S_y=X\setminus \{x\}\cup X\setminus \{y\}=X$. So $\mathcal{S}$ is a subbasis of $X$.

So, the basis associated with $\mathcal{S}$, $\mathcal{B}$, is the collection of subsets of $X$ with finite complement.

This is in fact a topology, which is the **finite complement topology** on $X$.

</details>
