conditions for a collection of subsets to be a basis for some topology

Not just any collection of subsets of $X$ can be a basis for a topology on $X$. For instance, if we took $𝒞$ to be all open intervals of length $1$ in $ℝ$, $𝒞$ isn’t the basis for any topology on $ℝ$: $(0,1)$ and $(.5,1.5)$ are unions of elements of $𝒞$, but their intersection $(.5,1)$ is not. The collection formed by arbitrary unions of members of $𝒞$ isn’t closed under finite intersections and isn’t a topology.

We’d like to know which collections $ℬ$ of subsets of $X$ could be the basis for some topology on $X$. Here’s the result:

Theorem.

A collection $ℬ$ of subsets of $X$ is a basis for some topology on $X$ if and only if:

1. Every $x\in X$ is contained in some $B_x\in ℬ$, and

2. If $B_1$ and $B_2$ are two elements of $ℬ$ containing $x\in X$, then there’s a third element $B_3$ of $ℬ$ such that $x\in B_3\subset B_1\cap B_2$.

Proof.

First, we’ll show that if $ℬ$ is the basis for some topology $𝒯$ on $X$, then it satisfies the two conditions listed.

$𝒯$ is a topology on $X$, so $X\in 𝒯$. Since $ℬ$ is a basis for $𝒯$, that means $X$ can be written as a union of members of $ℬ$: since every $x\in X$ is in this union, every $x\in X$ is contained in some member of $ℬ$. That takes care of the first condition.

For the second condition: if $B_1$ and $B_2$ are elements of $ℬ$, they’re also in $𝒯$. $𝒯$ is closed under intersection, so $B_1\cap B_2$ is open in $𝒯$. Then $B_1\cap B_2$ can be written as a union of members of $ℬ$, and any $x\in B_1\cap B_2$ is contained by some basis element in this union.

Second, we’ll show that if a collection $ℬ$ of subsets of $X$ satisfies the two conditions, then the collection $𝒯$ of unions of members of $ℬ$ is a topology on $X$.

• $\mathrm{\varnothing }\in 𝒯$: $\mathrm{\varnothing }$ is the null union of zero elements of $ℬ$.

• $X\in 𝒯$: by the first condition, every $X$ is contained in some member of $ℬ$. The union of all the members of $ℬ$ is then all of $X$.

• $𝒯$ is closed under arbitrary unions: Say we have a union of sets $T_{\alpha }\in 𝒯$…

  ${\displaystyle \bigcup _{\alpha \in I}}{T}_{\alpha }$$={\displaystyle \bigcup _{\alpha \in I}}{\displaystyle \bigcup _{\beta \in J_{\alpha }}}{B}_{\beta }$

  (since each $T_{\alpha }$ is a union of sets in $ℬ$)

  $={\displaystyle \bigcup _{\beta \in {\bigcup }_{\alpha \in I}{J}_{\alpha }}}{B}_{\beta }$

  Since that’s a union of elements of $ℬ$, it’s also a member of $𝒯$.

• $𝒯$ is closed under finite intersections: since a collection of sets is closed under finite intersections if and only if it is closed under pairwise intersections, we need only check that the intersection of two members $T_1,T_2$ of $𝒯$ is in $𝒯$.

  Any $x\in T_1\cap T_2$ is contained in some $B_x^1\subset T_1$ and $B_x^2\subset T_2$. By the second condition, $x\in B_x^1\cap B_x^2$ gets us a $B_x^3$ with $x\in B_x^3\subset B_x^1\cap B_x^2\subset T_1\cap T_2$. Then

  $$T_1\cap T_2=\bigcup _{x\in T_1\cap T_2}{B}_x^3$$

  which is in $𝒯$.

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