📂Topology

Equivalence Conditions for Bases in Topology

Definition ¹

A basis $\mathscr{B}$ for the topology $\mathscr{T}$ and a basis $\mathscr{B} '$ for the topology $\mathscr{T} '$, given in the set $X$, are considered to be equivalent if $\mathscr{T} = \mathscr{T} '$.

Theorem

The equivalence of bases is a necessary and sufficient condition for satisfying the following two conditions.

• (i): For all $B \in \mathscr{B}$ and $x \in B$, there exists $B ' \in \mathscr{B} '$ that satisfies $x \in B ' \subset B$.
• (ii): For all $B ' \in \mathscr{B} '$ and $x' \in b '$, there exists $B \in \mathscr{B}$ that satisfies $x' \in B \subset b '$.

Explanation

The concept of equivalent bases was developed as a way to express that, while the basis for a given topology may not be unique, they are essentially interchangeable. When viewing a basis as ‘ingredients to create a topology’, the condition $\mathscr{T} = \mathscr{T} '$ is a reasonable criterion for the equivalence of bases because if the resulting topology is the same, there is no meaningful distinction between them.

Proof

Let’s consider $\mathscr{B}$ and $\mathscr{B} '$ to be bases for $\mathscr{T}$ and $\mathscr{T} '$, respectively.

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$( \implies )$

Since $\mathscr{B}$ and $\mathscr{B} '$ are equivalent, $\mathscr{T} = \mathscr{T} '$ holds, and we can consider $B \in \mathscr{B}$ and $x \in B$.

In $$B \in \mathscr{B} \subset \mathscr{T} = \mathscr{T}’$$, because $\mathscr{B} '$ is a basis for $\mathscr{T} '$, $B$ is a union of some elements of $\mathscr{B} '$. Since $x \in B$, there exists $B ' \in \mathscr{B} '$ satisfying $x \in B ’ \subset B$, thus meeting condition (i), and by the same method, condition (ii) is also satisfied.

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$( \impliedby )$

Assuming (i) and (ii) hold, to prove $\mathscr{T} \subset \mathscr{T} '$, let’s say $U \in \mathscr{T}$ and $x \in U$.

Since $\mathscr{B}$ is a basis for $\mathscr{T}$, there exists $B_{x} \in \mathscr{B}$ satisfying $x \in B_{x} \subset U$. According to (i), for all $x$, there exists $B_{x} ' \in \mathscr{B} '$ satisfying $x \in B_{x} ' \subset B_{x}$, thus, $$U = \bigcup_{x \in U} B_{x}’$$ holds, proving $\mathscr{T} \subset \mathscr{T} '$. The same method can be applied to prove $\mathscr{T} ' \subset \mathscr{T}$, obtaining $\mathscr{T} = \mathscr{T} '$.

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