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

Not just any collection of subsets of

can be a basis for a topology on

. For instance, if we took $\mathcal{C}$ to be all open intervals of length

in $\mathbb{R}$ , $\mathcal{C}$ isn't the basis for any topology on $\mathbb{R}$ :

and

are unions of elements of $\mathcal{C}$ , but their intersection

is not. The collection formed by arbitrary unions of members of $\mathcal{C}$ isn't closed under finite intersections and isn't a topology.

We'd like to know which collections $\mathcal{B}$ of subsets of

could be the basis for some topology on

. Here's the result:

Proof. First, we'll show that if $\mathcal{B}$ is the basis for some topology $\mathcal{T}$ on

, then it satisfies the two conditions listed.

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

For the second condition: if and are elements of $\mathcal{B}$ , they're also in $\mathcal{T}$ . $\mathcal{T}$ is closed under intersection, so $B_1\cap B_2$ is open in $\mathcal{T}$ . Then $B_1\cap B_2$ can be written as a union of members of $\mathcal{B}$ , 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 $\mathcal{B}$ of subsets of satisfies the two conditions, then the collection $\mathcal{T}$ of unions of members of $\mathcal{B}$ is a topology on .

$\emptyset \in \mathcal{T}$ : $\emptyset$ is the null union of zero elements of $\mathcal{B}$ .
$X\in \mathcal{T}$ : by the first condition, every is contained in some member of $\mathcal{B}$ . The union of all the members of $\mathcal{B}$ is then all of .
$\mathcal{T}$ is closed under arbitrary unions: Say we have a union of sets $T_{\alpha}\in \mathcal{T}$ ...

$\displaystyle \bigcup_{\alpha \in I} T_{\alpha}$ $\displaystyle = \bigcup_{\alpha \in I} \bigcup_{\beta \in J_{\alpha}} B_{\beta}$

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

$\displaystyle = \bigcup_{\beta \in \bigcup_{\alpha \in I} J_{\alpha}} B_{\beta}$

Since that's a union of elements of $\mathcal{B}$ , it's also a member of $\mathcal{T}$ .
$\mathcal{T}$ 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 of $\mathcal{T}$ is in $\mathcal{T}$ .
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 with $x\in B_x^3 \subset B_x^1\cap B_x^2 \subset T_1\cap T_2$ . Then

$\displaystyle T_1\cap T_2 = \bigcup_{x\in T_1\cap T_2} B_x^3$

which is in $\mathcal{T}$ .

$\qedsymbol$