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..

  • $\begingroup$ 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 $F$? $\endgroup$
    – Asaf Karagila
    Jun 13 '19 at 9:34
  • $\begingroup$ F is a set of sets. also, I see what you mean, that answers my question. $\endgroup$
    – Qwertford
    Jun 13 '19 at 9:38
  • 1
    $\begingroup$ Since every finite set is bounded, wouldn't the same process show that every countable set is bounded? $\endgroup$
    – N. S.
    Jun 13 '19 at 9:43
  • $\begingroup$ 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 $\endgroup$
    – Asaf Karagila
    Jun 13 '19 at 9:46

The standard counterexample goes something like this.

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.


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