# Do countable $\cup$ and countable $\cap$ always exist?

Let $\Omega$ be a set and $(A_n)_{n\geq 1}$ a sequence of subsets of it. Does the countable union $\cup_{n=1}^{+\infty} A_n$ always exist? I understand that letting $B_n = A_1 \cup A_2 \cup \dots A_n$, $(B_n)_{n\geq 1}$ is an increasing sequence under the relation $\subseteq$, bounded above by $\Omega$.

Similarly, does the countable intersection of the $A_n$ sequence always exist since a similar $B_n$ is a decreasing sequence bounded by $\{\}$?

And from this can we conclude that $\cup_{n=1}^{\infty} \cap_{k=1}^{\infty} A_{n}^k$ always exists as a subset of $\Omega$?

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...What do you mean, always exist? $\bigcup_{i \in I} A_i$ is defined as the set of all elements $x \in \Omega$ for which there is at least one $i$ for which $x \in A_i$. Are you doing some set-theoretical stuff which makes you wonder about your own existence or something? In that case I could possibly feel your confusion. – Patrick Da Silva Jul 19 '13 at 0:41
But there are non-convergent sequences in $\mathbb{R}$ so isn't there also in $P(\Omega)$? I can imagine an always growing union of sets. – Enjoys Math Jul 19 '13 at 0:48

## 3 Answers

For any indexing set $I$ and any collection of subsets of $A_i\subset\Omega$ indexed by $I$, we have the following definitions:

$$\bigcup_{i\in I}A_i=\{a\in\Omega\mid \exists i\in I\text{ such that }a\in A_i\}\\\bigcap_{i\in I}A_i=\{a\in\Omega\mid a\in A_i\text{ for all }i\in I\}$$

Notice, we do not require anything of the cardinality of $I$. Using the above notation, if we have sets indexed by $I$ and $J$, say $A_i^j$, then we have:

$$\bigcup_{i\in I}\bigcap_{j\in J}A_i^j=\{a\in\Omega\mid \exists i\in I\text{ such that } a\in A_i^j\text{ for all }j\in J\}$$

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In ZF set theory, infinite unions exist because of a special axiom, the axiom of union, which states explicitly that if $\mathcal A$ is a set, then the union $$\bigcup{\mathcal A}\equiv\bigcup_{A\in\mathcal A} A$$ is a set.

For intersections we don't need a special axiom. Intersections exist because of a specification axiom, which says that for any set $X$ and any predicate formula $\Phi$, one can form the subset of all elements of $X$ for which $\Phi$ holds, written $$\{x\in X\mid \Phi(x)\}.$$ Then $\bigcap \mathcal A$ can be defined as $$\left\{ x\in \bigcup{\mathcal A}\mid \forall A\in{\mathcal A}.x\in A \right\}.$$ That is, as the set of all $x$ in the union of $\mathcal A$ such that $x$ is in $A$ for every set $A$ in the family $\mathcal A$.

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Yes, it always exists. It is defined as $\{\omega \in \Omega \mid \omega \in A_n, n\in \mathbb{Z}^+ \}$. The countable (or indeed, uncountable) intersection always exists and is defined similarly. Compounding these definitions, we can take the union of the intersections.

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