Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am curious why it's a problem to define a base using closed sets?

For example, my book uses the definition under "Constructing Topologies from Bases" as specified at, as opposed to the "definition" listed on this page. I don't see why closed intervals are a problem for example, the point ${1} \in [0,1], [1,2]$ so in particular $ {1} \in [0,1]\cap[1,2]=[1,1]=\{1\}$
I realize that topologies consist of "open sets" but why can't closed sets be a base for (larger) open sets for a topology.... or more generaly, why can't topologies be constructed using closed sets.

share|cite|improve this question
I'm confused. Say you have a "base" consisting of closed sets. What are the open sets in the topology you are defining with this "base"? – Alex Becker Dec 29 '12 at 1:42
No, my question is why not? You can always fit an open set in a larger closed set. And for example, the set of closed intervals covers $\mathbb{R}$ (which is both closed and open), so I don't see why they are not a base; unless we just require the base to consist of open sets (which is a common definition I've seen). – Squirtle Dec 29 '12 at 1:59
You might want to see this post – amWhy Dec 29 '12 at 2:01
Cool... thank you for this. I wasn't aware there was an equivalent definition: 1) The empty set and X are in τ . 2) The intersection of any collection of sets in τ is also in τ . 3) The union of any pair of sets in τ is also in τ . – Squirtle Dec 29 '12 at 2:05

$\newcommand{\ms}{\mathscr}$As is pointed out in the post linked from amWhy’s comment, one can construct a topology using closed sets. Recall that $\ms{T}\subseteq\wp(X)$ is a topology on $X$ iff

  • $\varnothing,X\in\ms{T}$;
  • $\bigcup\ms{U}\in\ms{T}$ whenever $\ms{U}\subseteq\ms{T}$; and
  • $U\cap V\in\ms{T}$ whenever $U,V\in\ms T$.

Suppose that $\ms T$ is a topology on $X$, and let $\ms C=\{X\setminus U:U\in\ms T\}$, the set of closed sets in $\langle X,\ms T\rangle$. Then it’s immediate from the De Morgan laws that $\ms C$ satisfies the following conditions:

  • $\varnothing,X\in\ms C$;
  • $\bigcap\ms F\in\ms C$ whenever $\ms F\subseteq\ms C$; and
  • $H\cup K\in\ms C$ whenever $H,K\in\ms C$.

It’s also clear that a family $\ms C\subseteq\wp(X)$ is the family of closed sets of some topology on $X$ iff $\ms C$ satisfies these conditions.

Next, recall that a family $\ms B\subseteq\wp(X)$ is a base for a topology on $X$ iff it satisfies the following conditions:

  • $\bigcup\ms B=X$, and
  • if $B_0,B_1\in\ms B$ and $x\in B_0\cap B_1$, then there is a $B_2\in\ms B$ such that $x\in B_2\subseteq B_0\cap B_1$.

In this case $\left\{\bigcup\ms U:\ms U\subseteq\ms B\right\}$ is a topology on $X$, and we say that $\ms B$ is a base for $\ms T$.

By looking at the complements of members of a base for a topology on $X$, we can see how the notion of a base for the closed sets ought to be defined. A family $\ms X\subseteq\wp(X)$ is a base for the closed sets of a topology on $X$ iff

  • $\bigcap\ms D=\varnothing$, and
  • if $D_0,D_1\in\ms D$ and $x\notin D_0\cup D_1$, then there is a $D_2\in\ms D$ such that $x\notin D_2\supseteq D_0\cup D_1$.

In this case $\left\{\bigcap\ms H:\ms H\subseteq\ms D\right\}$ is the family of closed sets of a topology on $X$, specifically, of the topology for which $\{X\setminus D:D\in\ms D\}$ is a base.

Now we can ask whether the closed intervals in $\Bbb R$ are a base for the closed sets of some topology on $\Bbb R$. They certainly cover $\Bbb R$. However, it’s not necessarily true that if $I_0$ and $I_1$ are closed intervals not containing $x$, then there is a closed interval $I_2$ such that $x\notin I_2\supseteq I_0\cup I_1$. For instance, let $I_0=[0,1]$, $I_1=[3,4]$, and $x=2$: any closed interval that contains $[0,1]\cup[3,4]$ necessarily also contains $2$. On the other hand, the collection of all subsets of $\Bbb R$ of the form $(\leftarrow,a]\cup[b,\to)$ with $a<b$ is a base for the closed sets of the usual topology on $\Bbb R$.

One can of course also ask whether the closed intervals of $\Bbb R$ are a base for the open sets of some topology on $\Bbb R$. In fact they are, but that topology isn’t the usual one: it’s an easy exercise to show that it’s the discrete topology. (HINT: If $a<b<c$, then $[a,b]\cap[b,c]=\{b\}$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.