# $S$ is closed under pairwise unions $⇒$ $S$ is closed under arbitrary unions?

Let $X$ be a set and $S$ be a collection of subsets of $X$, such that given any $U,V\in S$, $U\cup V\in S$.

Intuitively it seems like this should imply that arbitrary unions are also in $S$. That is, given index set $I$ and $\{U_i\}_{i\in I}\subseteq S$, $\bigcup_{i\in I}U_i\in S$.

Is this the case?

• NO. For example let $X$ be an infinite set and let $S$ be the collection of all finite subsets of $X.$ Let $A=\{\;\{p\}: p\in X\}.$ Then $A\subset S$ but $\cup A=X\not \in S.$ Also in this example, let $I=S$ and $U_i=i$ for each $i\in I.$ Then $\cup_{i\in I}U_i=\cup S=X\not \in S.$.... For a particular case let $S$ be the collection of all finite subsets of $\Bbb N$. Apr 16, 2018 at 16:53
• You can prove it by induction for arbitrary finite unions, but induction doesn't work anymore if you're working with infinite sets. Apr 16, 2018 at 18:37
• It’s false even for finite unions. In order to get arbitrary finite unions, you need closure under binary unions and under the nullary union (that is, $\varnothing\in S$). Apr 16, 2018 at 18:45
• @EmilJeřábek Could you please explain. I don't see yet, why you need the null case. May I suggest that you add an answer/counter example with a finite union (as far as I can tell all given counter examples so far use infinite index sets). Apr 17, 2018 at 7:31
• @MichaWiedenmann: This is mostly about what the term "finite union" means. By comparison, "finite sums" should logically allow for empty sums (since the empty set is certainly finite) whose value is $0$, so for instance the set $S=\{\,n\in\Bbb N\mid n>9\,\}$, though closed under pairwise sums, is not closed under arbitrary finite sums (the empty sum being the only culprit: we have $0\notin S$). For finite unions, it is similar. Apr 17, 2018 at 7:43

No, it's not the case. Let $S$ be the collection of finite subsets of $\mathbb{N}$, for example.

• Is this a counter-example if you allow only finite unions? Apr 16, 2018 at 23:23
Apr 16, 2018 at 23:33
• Your counter example seems to be that if you union all the finite subsets (an infinite number of them), you get $\mathbb{N}$ back, which is not a member of the set of finite subsets of $\mathbb{N}$. However, this requires unioning an infinite number of sets. What if you union only a finite number of sets from the collection? Is there still a counter example here? May 21, 2019 at 4:34

It is not the case. For example, the collection of (topologically) closed sets in $\Bbb R$ is closed under pairwise unions. However, the union $$(0,1) = \bigcup_{n = 1}^\infty [1/n,1-1/n]$$ is not a (topologically) closed set.

• Here we have closed used in two different senses, though still a correct answer Apr 17, 2018 at 0:15

You can also take $X=\mathbb{R}^2$ and let $S$ be the collection of bounded subsets. The union of two (or finitely many) bounded sets is bounded [prove it]. However, you can easily cover the entire plane with bounded "pieces" or "tiles" if you allow infinitely many of them [try that].

Consider the set $$\left\{\left[-1+\frac1n, 1-\frac1n\right]\mid n\in \Bbb N\right\}$$ of closed intervals on the number line.

• And again here we have closed used in a different sense, though still a correct answer Apr 17, 2018 at 0:14

Let $X = \mathbb{R}$, $S = \{(-\inf,x) \subset \mathbb{R} : x \in \mathbb{R} \}$. Observe that, for all $x,y \in \mathbb{R}$, $(-\infty,x) \cup (-\infty, y) = (-\infty, \max\{x,y\}) \in S$.

But $\bigcup_{i=0}^\infty (-\infty, i) = \mathbb{R} \not \in S$.