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

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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].

Answer 5

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