# Topology and Basis

## Definition (Topology)

A **topology** on a set $X$ is a collection $\mathfrak{T}$ having the following properties:
(1) $\emptyset$ and $X$ are in $\mathfrak{T}$
(2) The union of the elements of any subcollection of $\mathfrak{T}$ is in $\mathfrak{T}$. In other words,

(3) The intersection of the elements of any finite subcollection of $\mathfrak{T}$ is in $\mathfrak{T}$. In other words,

\[\begin{equation*} \text{If }U_1, \ldots, U_n \in \mathfrak{T}, \text{ then } \bigcap_{i=1}^n U_i \in \mathfrak{T}. \end{equation*}\]## Definition

Suppose that $\mathfrak{T}$ and $\mathfrak{T}^\prime$ are two topologies on a given set $X$. If $\mathfrak{T}^\prime \supset \mathfrak{T}$, we say that $\mathfrak{T}^\prime$ is **finer** than $\mathfrak{T}$; if $\mathfrak{T}^\prime$ properly contains $\mathfrak{T}$, we say that $\mathfrak{T}^\prime$ is **strictly finer** than $\mathfrak{T}$. We also say $\mathfrak{T}$ is **coarser** than $\mathfrak{T}^\prime$, or **strictly coarser**, in these two respective situations. We say $\mathfrak{T}$ is **comparable** with $\mathfrak{T}^\prime$ if either $\mathfrak{T}^\prime \subset \mathfrak{T}$ or $\mathfrak{T}^\prime \supset \mathfrak{T}$.

## Definition (Basis)

If $X$ is a set, a **basis** for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called **basis elements**) such that

(1) For each $x\in X$, there is at least one basis element $B_x\in\mathcal{B}$ such that $x\in B_x$.

(2) If $x$ belongs to the intersection of two basis elements $B_1, B_2 \in\mathcal{B}$, then there is a basis element $B_3\in\mathcal{B}$ such that $x\in B_3 \subset B_1\cap B_2$.

If $\mathcal{B}$ satisfies these two conditions, then we define the **topology $\mathfrak{T}$ generated by $\mathcal{B}$** as follows: A subset $U\subset X$ is open if for each $x\in U$, there is a basis element $B_x\in\mathcal{B}$ such that $x\in B_x\subset U$. In other words,

## Theorem 13.0

Let $\mathcal{B}$ be a basis for a topology on $X$. Then the **collection $\mathfrak{T}$ generated by $\mathcal{B}$** is a topology on X.

<*proof*>

If $U=\emptyset$, it satisfies the defining condition of openness vacuously. Similarly, $X\in \mathfrak{T}$ since for each $x\in X$ there is $B_x\in\mathcal{B}$ such that $x\in B_x$. Now, let us take an indexed family $\{U_\alpha\}_{\alpha \in \Lambda} \subset \mathfrak{T}$ and show that

\[\begin{equation*} U = \bigcup_{\alpha \in \Lambda}U_\alpha \in \mathfrak{T}. \end{equation*}\]Given $x\in U$, there is an index $\alpha$ such that $x\in U_\alpha$. Since $U_\alpha$ is open, there is a basis element $B\in\mathcal{B}$ such that $x\in B \subset U_\alpha$. Then $x\in B$ and $B\subset U$. Thus, $U$ is open.

Now let us take two elements $U_1, U_2 \in \mathfrak{T}$ and show that $U_1 \cap U_2 \in\mathfrak{T}$. Given $x\in U_1\cap U_2$, choose a basis element $B_1$ containing $x$ such that $B_1 \subset U$. Similarly, choose a basis element $B_2$ containing $x$ such that $B_2 \subset U_2$. The second condition for a basis enables us to choose a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$. Then $x \in B_3 \subset U_1 \cap U_2$, thus $U_1\cap U_2 \in \mathfrak{T}$.

Finally, we show by induction that any finite intersections $U_1 \cap \cdots \cap U_n$ of elements in $\mathfrak{T}$ is in $\mathfrak{T}$. This fact is trivial for $n=1$. Suppose it holds true for $n-1$ and prove it for $n$.

\[\begin{equation*} U_1 \cap \cdots \cap U_n = (U_1\cap \cdots \cap U_{n-1})\cap U_n \end{equation*}\]By the induction hypothesis, $U_1\cap\cdots \cap U_{n-1} \in \mathfrak{T}$. Since we have proved that intersection of two elements of $\mathfrak{T}$ is also in $\mathfrak{T}$. Thus, $U_1 \cap \cdots \cap U_n \in \mathfrak{T}$.

\[\tag*{$\square$}\]## Theorem 13.1

Let $X$ be a set and let $\mathcal{B}$ be a basis for a topology $\mathfrak{T}$ on $X$. Then $\mathfrak{T}$ equals to the collection of all unions of elements of $\mathcal{B}$.

<*proof*>
Given a collection of elements of $\mathcal{B}$, they are also elements of $\mathfrak{T}$. Because $\mathcal{T}$ is a topology, their union is in $\mathfrak{T}$. Conversely, given $U\in\mathfrak{T}$, choose for each $x\in U$ an element $B_x\in\mathcal{B}$ such that $x\in B_x\subset \mathcal{B}$. Then $U=\cup_{x\in U} B_x$, so $U$ equals a union of elements of $\mathcal{B}$.

## Lemma 13.2

Let $(X,\mathfrak{T})$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U \subset X$ and each $x\in U$, there is an element $C\in\mathcal{C}$ such that $x\in C\subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

<*proof*>
First, we need to show that $\mathcal{C}$ is a basis. Let $x\in X$ be given. Since $X$ is a open set itself, by the hypothesis, there is $C\in\mathcal{C}$ containing $x$. For the second condition of basis, let $x\in C_1\cap C_2$, where $C_1, C_2 \in \mathcal{C}$ be given. Since $C_1,C_2$ are open, so is $C_1\cap C_2$. Thus, by hypothesis, there is an element $C_3 \in \mathcal{C}\subset C_1\cap C_2$.

Let $\mathfrak{T}^\prime$ be the topology generated by $\mathcal{C}$. We want to show that $\mathfrak{T} = \mathfrak{T}^\prime$. Let an open set $U\in\mathfrak{T}$ be given. By the hypothesis, for each $x\in U$, there is $C\in\mathcal{C}$ such that $x\in C \subset U$. Then $U\in\mathfrak{T}^\prime$ by definition. Conversely, let $W\in \mathfrak{T}^\prime$ be given. Then

\[\begin{equation*} W = \bigcup_{x\in W}C_x\quad \text{where } x\in C_x \in \mathcal{C} \text{ and } C_x \subset W. \end{equation*}\]Since each $C_x \in \mathfrak{T}$ and $\mathfrak{T}$ is a topology, $\cup_{x\in W}C_x$ is open, i.e., $\bigcup_{x\in W}C_x\in\mathfrak{T}$.

\[\tag*{$\square$}\]## Theorem 13.3

Let $\mathcal{B}, \mathcal{B}^\prime$ be bases for the topologies $\mathfrak{T}$ and $\mathfrak{T}^\prime$, respectively, on $X$. Then the following are equivalent.

(1) $\mathfrak{T}^\prime \supset \mathfrak{T}$

(2) For each $x\in X$ and each basis element $B\in\mathcal{B}$ containing $x$, there is a basis element $B^\prime \in\mathcal{B}^\prime$ such that $x\in {B}^\prime \subset B$.

<*proof*>

(1) $\Rightarrow$ (2). Let $x\in X$ be given and let $B\in\mathcal{B}$ containing $x$ be given. Since $B \in \mathfrak{T}$ and $\mathfrak{T} \subset \mathfrak{T}^\prime$, $B\in\mathfrak{T}^\prime$. Since $\mathfrak{T}^\prime$ is generated by $\mathcal{B}^\prime$, there is a $B^\prime \in \mathcal{B}^\prime$ such that $x\in B^\prime \subset B$.

(1) $\Leftarrow$ (2). Let $U\in\mathfrak{T}$ be given. We want to show that $U \in \mathfrak{T}^\prime$. For each $x\in U$, there is $B\in\mathcal{B}$ such that $x\in B \subset U$. By the assumption, there is a basis element $B^\prime \in \mathcal{B}^\prime$ such that $x\in B^\prime \subset B$. Since $B\subset U$ and $\mathfrak{T}^\prime$ is a topology generated by $\mathcal{B}^\prime$, $U \in \mathfrak{T}^\prime$ by definition.

\[\tag*{$\square$}\]## Definition (Subbasis)

A **subbasis** $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The **topology generated by the subbasis** $\mathcal{S}$ is defined to be the collection $\mathfrak{T}$ of all unions of finite intersections of elements of $\mathcal{S}$.

<*proof*>
We want to show that $\mathfrak{T}$ is a topology. It suffices to show that collection $\mathcal{B}$ of all finite intersections of elements of $S$ is a basis. Let $x\in X$ be given. Then $x \in S$ for some $S\in \mathcal{S}$. Since $\mathcal{B}$ is unions of finite intersections of elements of $\mathcal{S}$, $S\in\mathcal{B}$. Now for second condition, let $x\in B_1 \cap B_2$, where $B_1, B_2 \in \mathcal{B}$, be given. Let

Then their intersection is \(\begin{equation*} B_1 \cap B_2 = (S_1 \cap \cdots \cap S_m) \cap (S^\prime_1 \cap \cdots \cap S^\prime_n) \end{equation*}\) also a finite intersection of elements of $\mathcal{S}$, so it belongs to $\mathcal{B}$.

\(\tag*{$\square$}\)

## Reference

- James Munkres,
**『**Topology**』**, Pearson