# (Un-)Countable union of open sets

Let $A_i$ be open subsets of $\Omega$. Then $A_0 \cap A_1$ and $A_0 \cup A_1$ are open sets as well.

Thereby follows, that also $\bigcap_{i=1}^N A_i$ and $\bigcup_{i=1}^N A_i$ are open sets.

My question is, does thereby follow that $\bigcap_{i \in \mathbb{N}} A_i$ and $\bigcup_{i \in \mathbb{N}} A_i$ are open sets as well?

And what about $\bigcap_{i \in I} A_i$ and $\bigcup_{i \in I} A_i$ for uncountabe $I$?

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Ok, it is an axiom for a topologies. But then there should be a proof for metrik spaces, say $\mathbb{R}$ with the canonical metrik? So to say, that every point $x \in \bigcap_{i \in I} A_i$ is an inner point. – Haatschii Jun 4 '12 at 20:16
A remark: regardless of whether it is true that an infinite union or intersection of open sets is open, when you have a property that holds for every finite collection of sets (in this case, the union or intersection of any finite collection of open sets is open) the validity of the property for an infinite collection doesn't follow from that. In other words, induction helps you prove a proposition for any natural number, but not for any transfinite cardinal. You'll have to use different techniques to prove or disprove the statement in your questions. – talmid Jun 4 '12 at 20:43

The union of any collection of open sets is open. Let $x \in \bigcup_{i \in I} A_i$, with $\{A_i\}_{i\in I}$ a collection of open sets. Then, $x$ is an interior point of some $A_k$ and there is an open ball with center $x$ contained in $A_k$, therefore contained in $\bigcup_{i \in I} A_i$, so this union is open. Others have given a counterexample for the infinite intersection of open sets, which isn't necessarily open. By de Morgan's laws, the intersection of any collection of closed sets is closed (try to prove this), but consider the union of $\{x\}_{x\in (0,1)}$, which is $(0,1)$, not closed. The union of an infinite collection of closed sets isn't necessarily closed.

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Any union of a set of open sets is again open. However, infinite intersections of open sets need not be open. For example, the intersection of intervals $(-1/n,1/n)$ on the real line (for positive integers $n$) is precisely the singleton $\{0\}$, which is not open.

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An arbitrary union (coutable or not) of open sets is open, but even for a countable intersection it's not true in general. For example, when $\Omega$ is the real line endowed with the usual topology, and $A_i:=\left(-\frac 1i,\frac 1i\right)$, $A_i$ is open but $\bigcap_{i\in \Bbb N}A_i=\{0\}$ which is not open.

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Moreover (exercise for the reader), any subset of the ambient space (e.g. the reals) can be written as the intersection of some collection of open sets. – Dave L. Renfro Jun 4 '12 at 21:51
@DaveL.Renfro: can you give some hints how to prove this? – T. Eskin Jun 5 '12 at 11:28
Actually, Dave's assertion fails for the indiscrete topology... – GEdgar Jun 5 '12 at 12:16
@Thomas E.: Assume singleton sets are closed sets (to take care of GEdgar's observation). Then, given any subset $E,$ we can write $E$ as the union of a collection of singleton sets (use the points belonging to $E$), and hence $E$ can be written as the union of a collection of closed sets. By applying De Morgan's Law, we can now write the complement of $E$ as the intersection of a collection of open sets (the open sets will be co-singleton sets). Finally, note that as $E$ varies over all subsets of the ambient space, the complement of $E$ will vary over all subsets of the ambient space. – Dave L. Renfro Jun 5 '12 at 16:25
@DaveL.Renfro. True, I didn't somehow even think about that :-) So $T_{1}$ would be the least requirement for this construction. – T. Eskin Jun 5 '12 at 19:32