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In a question I'm working on it asked for an example of an infinite intersection of open sets that itself is open.
Just wondering – let's say that I were to take ${\mathbb R}^{2}$ with the usual metric defined on it. Then say the intersection the open balls centered on some arbitrary $x$ of radius less than $r$ such that $r>1$. Would this constitute a suitable answer? Thanks in advance!
No, your intersection would be the closed ball of radius $1$, which is not open.
You can just take the "constant family" $A_i=A$ for all $i$ in some (infinite) set of indices and $A$ an open set. If you want something non-constant, consider $A_n=(n,\infty)$ in $\mathbb{R}$.
If you can choose an arbitrary topology for your example, one easy choice would be to choose the discrete topology (every set is open) on some arbitrary set, and then use any collection of sets. Since the intersection of sets is always a set, and in the discrete topology all sets are open, you are guaranteed to get an open set.
Another construction that works in any topological space is to select some open set, and then consider a collection consisting of that set and open supersets of that set. Then the intersection of all those sets is exactly the set you started with, which is open.
Yet another possibility is to use any collection of open sets that contains two disjoint sets. Then the intersection of those sets will be empty, and the empty set is open in any topology.
Indeed, while your example is wrong, with a slight change it can be made into an construction of the second type: Instead of considering the open balls around $x$ with $r>1$, consider $r\ge 1$. Then all the open balls are supersets of the open ball with $r=1$, which is in the set, and therefore is the intersection of all those balls.
