Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Omega$?

share|improve this question
4  
...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 3

up vote 3 down vote accepted

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:

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

share|improve this answer

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.

share|improve this answer

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 $\mathcal A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.