Math4201 Topology I (Lecture 4)

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 $x\in C_3\subseteq U=C_1\cap C_2$.

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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 )

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

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 $\mathbb{R}$.

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 $X$.