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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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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Kasper Apr 26 '13 at 19:37
    
@Kasper Thanks! I'll be sure to add more context next time. –  john.abraham Apr 27 '13 at 9:20

1 Answer 1

up vote 1 down vote accepted

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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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? –  john.abraham Apr 27 '13 at 10:18
    
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. –  Asaf Karagila Apr 27 '13 at 10:20

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