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Since union of sets is defined to be independent of the number of sets as in Wikipedia

If M is a set whose elements are themselves sets, then x is an element of the union of M if and only if for at least one element A of M, x is an element of A.

I was wondering why for a class of subsets of a set, it being closed under finite union and it being closed under countable union are not same?

Similar question for intersection.

Thanks and regards!

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up vote 3 down vote accepted

Starting from a initial collection of sets being allowed to take countable unions and intersections lets you create many more sets that being allowed to take only finite unions and intersections. Therefore it seems plausible to me that the former can take you out of your starting collection even if the latter does not.

You can find examples in any standard measure theory text. Here is a very simple one. Let $$A_n=\{1,2,3,\ldots,n\}$$

Let $\Omega$ be the collection of the sets $A_n$ for $n \in \mathbb{N}$.

Then for a finite set of indices, $n_1,n_2,\ldots,n_k$ say, the union $$A_{n_1} \cup A_{n_2} \cup \cdots \cup A_{n_k} = A_{\max\{n_1,\ldots,n_k\}}$$ and hence belongs to $\Omega$. So $\Omega$ is closed under finite unions.

But the countably infinite union $$\cup_{n=1}^{\infty} A_n = \{1,2,3,\ldots\}$$ and hence does not belong to $\Omega$. So $\Omega$ is not closed under countable unions.

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Hmm. There is a somewhat sneaky and quite false conclusion being made here. It's true that the Wikipedia definition doesn't mention the number of sets. However, that doesn't mean that union and intersection are "independent of" the number of sets. I can't even figure out what this would mean. To state an obvious counterexample, adding another set to your union will give a set with at least as many elements as before, and adding another set to your intersection will give a set with at most as many elements as before. (Not even in the sense of cardinality, but just the simple fact that $$\left(\bigcup_\alpha A_\alpha\right)\cup B\supset \bigcup_\alpha A_\alpha$$ and the dual statement for intersections.

Jyotirmoy's example shows that the countable union of finite sets isn't finite. Another one comes from topology. A topology on a set $X$ is defined to be a set of subsets of $X$ that is closed under unions and finite intersections. We call such sets "open," and they're roughly a way of saying what points are close to each other. A typical example is the Euclidean topology on $\mathbb{R}$, which consists of the open intervals $(a,b)$ and their unions. This is well-defined because any finite intersection of open intervals is either the empty set or another, smaller open interval. On the other hand, the set $\bigcap_n (-1/n,1/n)$ is just the point $\lbrace 0\rbrace$, which is definitely not a union of open intervals.

On the other hand, countable intersections are still kind of important in the same general situation. A $\sigma$-algebra is a set of subsets of $X$ that is closed under complementation and countable intersections (and thus also countable unions). You can use such a thing to define a "measure," which is an assignment of a sort of "size" to each set in the $\sigma$-algebra. The algebra of Lebesgue-measurable on $\mathbb{R}^n$ is probably the best example, but you can turn any topology into a $\sigma$-algebra, called its Borel algebra, just by iterating countable intersections and countable unions. A set that's a countable intersection of open sets is called a $G_\delta$, from the German words "gebiet Durchschnitt" meaning "open intersection." So we know, for example, that every $G_\delta$ in $\mathbb{R}$, though not necessarily open, has a Lebesgue measure.

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