# 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 $\mathcal{C}$ to be all open intervals of length $1$ in $\mathbb{R}$, $\mathcal{C}$ isn’t the basis for any topology on $\mathbb{R}$: $(0,1)$ and $(.5,1.5)$ are unions of elements of $\mathcal{C}$, but their intersection $(.5,1)$ 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 $X$ could be the basis for some topology on $X$. Here’s the result:

###### Theorem.

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

1. 1.

Every $x\in X$ is contained in some $B_{x}\in\mathcal{B}$, and

2. 2.

If $B_{1}$ and $B_{2}$ are two elements of $\mathcal{B}$ containing $x\in X$, then there’s a third element $B_{3}$ of $\mathcal{B}$ such that $x\in B_{3}\subset B_{1}\cap B_{2}$.

###### Proof.

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

$\mathcal{T}$ is a topology on $X$, so $X\in\mathcal{T}$. Since $\mathcal{B}$ is a basis for $\mathcal{T}$, that means $X$ 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 $B_{1}$ and $B_{2}$ 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 $X$ satisfies the two conditions, then the collection $\mathcal{T}$ of unions of members of $\mathcal{B}$ is a topology on $X$.

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

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

• $\mathcal{T}$ 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 $T_{1},T_{2}$ 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 $B_{x}^{3}$ 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 $\mathcal{T}$.

Title conditions for a collection of subsets to be a basis for some topology ConditionsForACollectionOfSubsetsToBeABasisForSomeTopology 2013-03-22 14:21:49 2013-03-22 14:21:49 waj (4416) waj (4416) 4 waj (4416) Proof msc 54A99 msc 54D70