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We all know the definition of a topology $T$ on a nonempty set $X$: A collection $T$ of subsets of a nonempty set $X$ is a topology on $X$ if, (1) $\emptyset,X\in T$ (2) For any arbitrary collection of elements of $T$ their union is also in $T$ (3) For any finite collection of elements of $T$ their intersection is also in $T$.

My question is why in the definition only the finite intersection is allowed why not arbitrary intersection?

metamorphy
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Shreya Chauhan
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    Consider $\bigcap_{n=1}^\infty(-1/n,1/n)$. – Angina Seng Sep 24 '20 at 02:25
  • Openness is extended from usual metric in R. Thick about Seng's saying. – Han Sep 24 '20 at 02:29
  • For $\mathbb R$, starting with open intervals you can get any singleton as an intersection of open intervals (as in @AnginaSeng's example). And then you get every subset as a union of singletons. Similarly for any metric space, starting with open balls. – Robert Israel Sep 24 '20 at 03:07

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Let $X$ be a Hausdorff topological space (that means any two distinct point $x,y \in X$ are contained in disjoint open sets $U_x,U_y$ respectively -- most interesting topological spaces, such as Euclidean spaces, are Hausdorff). Suppose $X$ satisfies the property that arbitrary intersections of open sets are open. Let $z \in X$ and let $I_z$ be the intersection of all open sets that contain $z$. If $y \neq z$ then there is an open set $U_z$ containing $z$ but not $y$, so $y \not\in I_z$. Thus $I_z = \{z\}$, so all points are open. But if all points are open, that means by arbitrary union that every single last subset of $X$ is open, in which case why did we bother going through the trouble of defining a topology in the first place??

In summary, allowing arbitrary intersections of open sets to be open implies that any Hausdorff space is discrete, which basically kills the entire field of topology... so I think sticking with finite intersections is the way to go.

Al Suarizmi
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