Base

Definitiom: Base for a Topological Space

Let $(X,\tau )$ be a topological space.

A base for $(X,\tau )$ is a collection $\mathcal{B}$ of open subsets of $X$ such that every open set can be represented as a union of a subset of $\mathcal{B}$.

WARNING

This representation is not necessarily unique.

Base Criteria

Theorem

Let $(X,\tau )$ be a topological space.

A collection $\mathcal{B}$ of open sets is a base for $(X,\tau )$ if and only if for each open set $U$ and each $u\in U$, there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$.

PROOF

We need to prove two things:

• (I) If $\mathcal{B}$ is a base for $(X,\tau )$, then for each open set $U$ and each $u\in U$, there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$.
• (II) If for each open set $U$ and each $u\in U$, there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$, then $\mathcal{B}$ is a base for $(X,\tau )$.

Part I:

Suppose $\mathcal{B}$ is a base for $(X,\tau )$. Let $U$ be open and let $u\in U$. Since $\mathcal{B}$ is a base, $U$ can be represented as a union of some subset ${\mathcal{B}}_U\subseteq \mathcal{B}$, i.e. $U=\bigcup {\mathcal{B}}_U$. This implies that every $B\in {\mathcal{B}}_U$ is a subset of $U$. Moreover, since $u\in U$, there must exist at least one $B\in {\mathcal{B}}_U$ which contains $u$.

Part II:

Suppose that for each open set $U\in \tau$ and each $u\in U$, there exists a $B(u)\in \mathcal{B}$ such that $B(u)\subseteq U$ and $u\in B(u)$. Since $B(u)\subseteq U$ for each $u\in U$, it holds that ${\bigcup }_{u\in U}B(u)\subseteq U$. Since this union contains every $u\in U$, it follows that ${\bigcup }_{u\in U}B(u)=U$. Therefore, $U$ is a union of a subset of $\mathcal{B}$, Q.E.D.

Topology Generation

Theorem: Topology Generation

Let $(X,\tau )$ be a topological space and let $\mathcal{B}$ be a base for $(X,\tau )$.

A subset $U\subseteq X$ is open if and only if for each $u\in U$ there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$.

PROOF

We need to prove two things separately:

• (I) If $U$ is open, then for each $u\in U$ there exists some $B_u\in \mathcal{B}$ such that $B_u\subseteq U$ and $u\in B_u$.
• (II) If for each $u\in U$ there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$, then $U$ is open.

Proof of (I):

Suppose $U$ is open. The fact that for each $u\in U$ there exists some $B_u\in \mathcal{B}$ such that $B_u\subseteq U$ and $u\in B_u$ follows immediately from the base criterion.

Proof of (II):

Suppose that for each $u\in U$ there exists some $B\in \mathcal{B}$ such that $B\subseteq U$ and $u\in B$. Since $u$ is contained in $B_u$ and $B_u\subseteq U$, we know that $U={\bigcup }_{u\in U}{B}_u$, i.e. $U$ is a union of open subsets and is thus open.

INTUITION

This theorems allows us to determine the topology $\tau$ given only a base for it.