# 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.

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mathoverflow.net/questions/19152/… – k.stm Jan 23 '13 at 12:10
Check out mathoverflow.net/questions/19152/… for a lot of good insights. – Grumpy Parsnip Jan 23 '13 at 12:11
@Jim Yeah, I already covered that. :sunglasses: – k.stm Jan 23 '13 at 12:12
@K.Stm. Your comment wasn't yet visible when I posted mine. :) – Grumpy Parsnip Jan 23 '13 at 12:23
Thanks for the links ! – Alan Simonin Jan 23 '13 at 12:54

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.