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Let's say I have a non-empty set $X$ and a collection $\mathcal{A}$ of subsets of $X$. And let's say I have $ \bigcup_{n\in\mathbb{N}}A_n$ where $A_n\in\mathcal{A}$.

Question 1:

If $\emptyset,\lbrace 1,2\rbrace \in \mathcal{A}$. Does $ \bigcup_{n\in\mathbb{N}}A_n$ to refer to

  • $(\emptyset\cup \emptyset\cup\dots\cup \emptyset\cup \dots )$

  • $(\lbrace 1,2\rbrace\cup \lbrace 1,2\rbrace\cup\dots\cup \lbrace 1,2\rbrace\cup \dots )$

  • $(\emptyset\cup \lbrace 1,2\rbrace\cup\dots\cup \emptyset\cup \lbrace 1,2\rbrace\cup \dots )$

or any other arbitrary union of sets from $\mathcal{A}$?

Question 2:

If $ \bigcup_{n\in\mathbb{N}}A_n\notin \mathcal{A}$, does that mean that $\mathcal{A}=\emptyset$? If $\mathcal{A}$ contained a set, couldn't I then create an infinite union of that set?

Question 3:

Can I take any collection of sets $\mathcal{A}$ and then say that $ \bigcup_{n\in\mathbb{N}}A_n= \mathcal{A}$ where $A_n$ is then all the subsets of $\mathcal{A}$?

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When we write $A_n\in\cal A$ and $\bigcup_{n\in\Bbb N}A_n$ then we talk about a specific sequence of sets, whether or not these are $\varnothing$ or $\{1,2\}$ depends on the situation, the context and its generality.

But the general rule of thumb is that $\bigcup_{n\in\Bbb N}A_n=A_0\cup A_1\cup\ldots=\{x\in X\mid\exists n\in\Bbb N: x\in A_n\}$.

If that union is not in $\cal A$ it means just that $\cal A$ is not closed under countable unions. Consider the case where $X=\Bbb N$ and $\cal A$ is all the singletons, and let $A_n=\{n\}$. Then the union of the $A_n$'s is certainly not a singleton and therefore not in $\cal A$.

Your last question seems a bit out of context. Is $\cal A$ a set of subsets of $X$ or subsets of itself? It is certainly possible to have $\cal A$ as the countable union $\bigcup_{n\in\Bbb N}A_n$, but then we doesn't always have $A_n\in\cal A$, rather we have $A_n\subseteq\cal A$.

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  • $\begingroup$ Thanks so much for your reply. A quick question if you don't mind: If we have two collections of sets $\mathcal{A},\mathcal{B}$, the set $\lbrace A \cup B : A \in \mathcal{A} \,and\, b \in \mathcal{B} \rbrace$ contains all combinations unions of $A$ and $B$, right? $\endgroup$ – john.abraham Apr 27 '13 at 10:18
  • $\begingroup$ john, yes. This would be all the possible unions of a pair $A,B$ from $\cal A,B$ respectively. Note however that it contains only these unions of pairs, and not arbitrary unions from $\cal A$ and $\cal B$. It is perfectly possible that either collection is not closed under finite unions, or infinite unions, in which case this is not sufficient to conclude that we have covered all the unions. $\endgroup$ – Asaf Karagila Apr 27 '13 at 10:20

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