The set of all closed intervals of the real line also satisfies the open sets axioms.

1.The empty set and X itself are open.

2.Any union of open sets is open.

3.The intersection of any finite number of open sets is open.

Why the standard real line topology is defined on open intervals rather than using closed intervals as open sets?

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    $\begingroup$ What axioms are you talking about? That the set of open sets is closed under finite intersections and arbitrary unions? What is the union of all closed intervals of the form $[-1+1/n,1-1/n]$ for $n\in \Bbb N, n\geq 2$? Is that closed? $\endgroup$ – Arthur May 31 '15 at 20:03
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    $\begingroup$ The set of closed intervals does not satisfy the open set axioms, but rather a dual version of them. Namely, arbitrary intersection of closed sets is closed, and finite union of closed sets is closed. $\endgroup$ – Pedro M. May 31 '15 at 20:04
  • $\begingroup$ @Arthur and Pedro, you are right. If any of these comments move to answer, I will accept it.thanks. $\endgroup$ – ahala May 31 '15 at 20:14

Another point of view...

For any set $\mathcal{S}$ of subsets of $\mathbb{R}$, there is a smallest topology $\mathcal{T}$ on $\mathbb{R}$ such that $\mathcal{S} \subseteq \mathcal{T}$. See here for details. So, in principle, you can use any set of subsets of $\mathbb{R}$ to generate a topology. However, the topology obtained will usually be different from the "standard topology" on $\mathbb{R}$, i.e. the one generated by the set of all open intervals.

In particular, let's see what topology $\mathcal{T}$ is generated by the set of all closed intervals of $\mathbb{R}$. For any $x \in \mathbb{R}$ we know $[x,x+1]$ and $[x-1,x]$ are closed intervals, so $[x,x+1],[x-1,x] \in \mathcal{T}$. However, $\mathcal{T}$ is a topology, so it is closed under finite intersections. Thus $\{x\} = [x,x+1] \cap [x-1,x] \in \mathcal{T}$. So, $\mathcal{T}$ contains every singleton $\{x\}$ where $x \in \mathbb{R}$. Now, using that $\mathcal{T}$ is closed under arbitrary unions, you should be able to see that every subset of $\mathbb{R}$ belongs to $\mathcal{T}$. That is, the topology $\mathcal{T}$ on $\mathbb{R}$ generated by the set of closed intervals is the entire power set of $\mathbb{R}$ -- also known, in this context, as the discrete topology on $\mathbb{R}$.

So, in conclusion, using the closed intervals to generate a topology on $\mathbb{R}$ is not exactly "forbidden", but it does lead to a topology on $\mathbb{R}$ which is drastically different from (and much less useful than) the standard topology.

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As noted in the comment of Arthur, intervals on the real line of the form $a\le x\le b$ does not , in general, satsfies the axiom 2), so they does not form a family of set that satisfies the three axioms that define a family of open set.

But a topology can be defined also by a family of closed set (see here) using the fact that closed sets satisfy a dual version of the axioms defining the open sets, as noted by Pedro M..

The topological spaces generated on the real line by the family of open intervals (using axioms for open sets) or by the family of closed intervals (using axioms for closed sets) are the same.

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