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.
$\newcommand{\ms}{\mathscr}$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 $a&lt;b$ 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 $a&lt;b&lt;c$, then $[a,b]\cap[b,c]=\{b\}$.)
The simple answer is that if you wanted closed intervals to form a basis of some topology, then you would, for example, have to allow ${2}$ (which is the intersection of $[1,2]$ and $[2,3]$) to be a closed interval itself. But look at the question in your link. It says:
Show that the collection $\mathcal{C}=\{[a,b]:a,b\in\mathbb{R},a&lt;b\}$ of all closed intervals in $\mathbb{R}$ is '''not''' a base for a topology on $\mathbb{R}$.
It categorically says that [2,2] is not a valid basis element. So, I guess that should clear it out.
However, as a fun fact, if you were to allow singletons, your topology is just the discrete topology. Cheers!