# Union is finite implies the collection is finite

Let $C$ be a collection of sets, and $\bigcup C \in V_{\omega}$ where $V_{\omega}$ is the collection of hereditarily finite sets. Is it possible to show that $C\in V_{\omega}$?

YES. Because $V_\omega$ models every axiom of set theory except the infinity axiom;

i.e. $\bigcup C \in V_{\omega}\:\Rightarrow\:C \subseteq \mathcal{P}\left(\bigcup C\right)\in V_{\omega}\:\Rightarrow\: C\in \mathcal{P}\left(\mathcal{P}\left(\bigcup C\right)\right)\in V_{\omega}\:\Rightarrow\: C\in V_{\omega}$.

QUESTION What if I instead assume $\bigcup C$ is simply finite. Can I show that $C$ is finite?

Initial Thoughts From the assumption, it follows that every element of $C$ is finite. Is it possible to have an infinite collection of finite sets whose union is finite?

EDIT: $C\subseteq \underbrace{\mathcal{P}(\bigcup C)}_{\text{finite}}$. "Thank you" to a recently deleted answer which brought my senses back.

• If by "collection of sets" you mean a family $(C_\iota)_{\iota\in I}$ with an index set $I$ then you could have $C_\iota=\{1\}$ for all $\iota\in I$. – Or have I missed something? – Christian Blatter Feb 9 '16 at 14:52
• "families" = "functions" in set theory. By $\{ C_{i} \}_{i\in I}$ you mean to say the image of $C$ (the function) with respect to $I$. If $C$ is constant, then $\{ C_{i} \}_{i\in I}$ is a singleton. – Alberto Takase Feb 9 '16 at 15:35

## 1 Answer

Sorry, I accidentally deleted my response.

It's not possible for the union to be finite and the collection infinitie. The point here is that, we can note that if $U=\cup C$ then $C\subseteq P(U)$. The power set of S is finite if and only if S is finite..

• @skyking Try rereading the question – Stella Biderman Feb 9 '16 at 14:31
• Ah, you're right. The wording of the question changed a bit. I'll reword my answer – Stella Biderman Feb 9 '16 at 14:46