# Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections?

As in the title, I am asking if there is a difference between allowing set-theoretic operations over arbitrarily many sets, and restricting to only countably many sets.

For example, the standard definition of an topology on a set $X$ requires that arbitrary unions of open sets are open. Do I lose anything significant if I restrict this to just unions of countably many (open) sets?

I cannot come up with an example where it makes a difference.

• Take the disjoint union of uncountably many copies of the real line. Then the whole space cannot be expressed as the countable union of open balls. (Of course, the space isn't a metric space either, but it is locally a metric space in some sense...) – Zhen Lin Jun 25 '12 at 15:03
• More generally, take any space without a countable basis. Then there will be some open sets you will not be able to obtain as a countable union of basic open sets. E.g. any discrete (so metric!) uncountable space. – tomasz Jun 25 '12 at 15:04
• @Zhen: sure it's a metric space; you just need to allow the metric to take the value $\infty$. (If you want coproducts in the category of metric spaces you should do this anyway.) – Qiaochu Yuan Jun 27 '12 at 4:10

Let $X$ be an uncountable set. Let $\tau = \{ O \subseteq X \mid O = X \text{ or } O \text{ is at most countable}\}$ Then $\tau$ contains $\emptyset$ and $X$, is closed under finite intersections and under countable unions. But it isn't a topology on $X$ as it isn't closed under arbitrary unions. So it makes a difference.

• Can you please explain why $\tau$ is closed under countable unions? Because since $\tau$ only contain $\emptyset$ and $X$(which is a finite set), shouldn't it be closed under finite union? – Wasiq Noor Aug 12 '18 at 11:09

Yes, it makes a difference. Here’s a very simple example. Let $D$ be any uncountable set; take it to be $\Bbb R$, if you want to be very specific. Let $\mathscr{A}=\{X\}\cup\{A\subseteq D:A\text{ is countable}\}$. The family $\mathscr{A}$ is closed under finite intersections and countable unions, but it’s not a topology, because it’s not closed under arbitrary unions. In particular, if $F$ is any non-empty finite subset of $D$, then $D\setminus F\notin\mathscr{A}$. However, the topology on $D$ generated by $\mathscr{A}$ is the discrete topology, in which every subset of $D$ is open, since every subset of $D$ is the union of countable (indeed finite) subsets of $D$: if $A\subseteq D$, then $$A=\bigcup_{x\in A}\{x\}\;.$$

*Edit:*I originally let $X$ be just a Hausdorff space, but in that case the two aren't guaranteed to be different.

Here's an example to see a where it makes a difference. Let $X$ be an uncountable Polish space. If you look at the smallest collection of subsets of $X$ that a) contains all the open sets of $X$ and b) is closed under complements and countable unions (hence also countable intersections), you get the Borel $\sigma$-algebra of $X$.

But if you look at the smallest collection of subsets of $X$ that a) contains all the open sets and b) is closed under complements and arbitrary unions (hence also arbitrary intersections), this is $\cal{P}$$(X)$. This is because it includes all closed sets, hence all singletons, and then we can take an arbitrary union to get any subset of $X$.

• I believe this example has the advantage of showing that there's a difference even in cases where X is countably infinite. The uncountability of X in other examples is therefore not required for the proof. – MikeC Jun 26 '12 at 22:06
• It is not obvious that the Borel $\sigma$-algebra can be different from $\mathcal{P}(X)$. For example, it is consistent with ZF that the Lebesgue $\sigma$-algebra of $\mathbb{R}$ is $\mathcal{P}(\mathbb{R})$. So you have not actually shown that this distinction matters (that is you have not exhibited a non-Borel set). – Qiaochu Yuan Jun 27 '12 at 4:05
• @QiaochuYuan You are right: for any countable discrete space, they are the same. But if we assume $X$ is uncountable Polish (separable and completely metrizable), then there is an analytic non-Borel set. We could maybe get away with less, but we certainly need more than just being Hausdorff. – Francis Adams Jun 27 '12 at 13:02