In a topological space, why the intersection only has to be finite? I was wondering if there is a reason why in the definition of a topologic space, the reunion of any collection of open sets is also an open set but the intersection has to be finite to be a set.
Is there because of the definition of an open set itself ? I have always seen it as a definition, but now I want to know why.
Thanks for your help.
 A: This is an alternative approach to the definition of topology, thinking of topology in general as representing intuitionist logic. This by no means should be your primary argument, I just always liked this notion.
You can think of open sets as sets for which we can have "finite witnesses." That is, we can show a point $x$ is in the open set $U$ with a finite amount of information. 
Then if $\mathcal U$ is a set of open subsets of a topological space, to show that $x\in\bigcup \mathcal U,$ we need only finite information that $U\in\mathcal U$ and finite information showing that $x\in U.$
On the other hand, if $\mathcal U$ has infinitely many sets, to show that $x\in\bigcap \mathcal U,$ we'd have to have an infinite list of finite information showing $x\in U$ for each $U\in\mathcal U.$ So there is no reason to think we can provide for such a general $x$ a finite proof that $x\in\bigcap \mathcal U.$
A: This is an axiom of Topological spaces. This was defined so, because if you allow for infinite intersections of open sets, you get undesirable sets that are "open". One such example would be the intersection of all sets of the form:
$$\bigcap_{n \in \mathbb{Z}} \left(-\frac{1}{n},\frac{1}{n}\right)$$ 
in the standard metric space defined over $\mathbb{R}$. It is clear that this "open" set is not open in the equivalent definition for metric spaces - which demand that a neighborhood of the point be inside the open set. Thus, if you want the metric space definition of a open set to coincide with the topological one, you must define axiomatically that infinite intersection are not necessarily open. 
