Is every element a union of some subset, for a set closed under arbitrary unions?

Question

Consider a set $A$ satisfying $Y\subset A\implies \bigcup Y\in A$ for any $Y$.
Now, I intuitively expect that any set $X\in A$ will satisfy $X = \bigcup Y$ for some $Y\subset A$. But I am not completely sure, and I have no idea how to prove or disprove it.

Any help?

@GeoffreyTrang Sorry, I don't get you example/counterexample. — Atom, Apr 01 '21 at 20:36
You've gotten a good Answer, but you might want to note that the union over an empty family of sets is by convention the empty set. — hardmath, Nov 17 '21 at 00:44

score 1 · Accepted Answer · edited Nov 18 '21 at 15:20

1

If $X\in A$, then $X\subset A$
whence $X = \bigcup \{X\}$.

edited Nov 18 '21 at 15:20

Atom

3,905

answered Apr 01 '21 at 21:29

William Elliot

17,598

Oh! How could I not get this! – Atom Apr 01 '21 at 21:42

Is every element a union of some subset, for a set closed under arbitrary unions?

1 Answers1