# Base (topology) with closed intervals

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 http://en.wikibooks.org/wiki/Topology/Bases, 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.

• 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 – Namaste 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 $[a,b]\cap[b,c]=\{b\}$.)