# how is closed under finite union not the same as closed under countable union?

How is closed under finite union not the same as closed under countable union?

I know that closed under countable union should be a stronger requirement than closed under finite union. So why does the following argument not work?

Let {$$A_1,A_2,...$$} be a countably infinite sequence of sets in F. Then if F is closed under countable union, the countable union of {$$A_1,A_2,...$$} should also be in F.

However, If we assume F was closed only under finite union, couldn't you use induction to show that the countable union of {$$A_1,A_2,...$$} will also be in F? That is $$A_1 \bigcup A_2$$ is in F, $$A_1 \bigcup A_2 \bigcup A_3$$ in in F and so on..

• Finite unions of finite sets are finite. Countable unions of finite sets don't have to be. But generally speaking, it's good to put context in your question. What's $F$? Jun 13 '19 at 9:34
• F is a set of sets. also, I see what you mean, that answers my question. Jun 13 '19 at 9:38
• Since every finite set is bounded, wouldn't the same process show that every countable set is bounded? Jun 13 '19 at 9:43
• Let me link to a few threads that might be enlightening, given your suggestion to use induction. math.stackexchange.com/questions/1437036 math.stackexchange.com/questions/307277 math.stackexchange.com/questions/717961 Jun 13 '19 at 9:46

Take $$A_n = \{0,1,\dots,n\}$$, and let $$S$$ be the collections of all sets $$A_n$$. Then we have that finite unions, say $$\bigcup_k A_{n_k}$$, that this belongs to $$S$$, just by considering that the $$A_{\max {n_k}}$$ will be a element of $$S$$.
But consider the countable union $$\bigcup_{n \in \mathbb{N}} A_n = \{1, 2, \dots\}$$. This is not an element of $$S$$, which can be shown by a simple argument from contradiction, and follows from the fact that this is no longer a finite set.