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Are there necessary conditions to be able to switch infinite set operations?

Or is it always true that:

$\displaystyle \bigcup_{i=1}^\infty \bigcup_{j=1}^\infty A_{ij} = \bigcup_{j=1}^\infty \bigcup_{i=1}^\infty A_{ij} $

$\displaystyle \bigcap_{i=1}^\infty \bigcap_{j=1}^\infty A_{ij} = \bigcap_{j=1}^\infty \bigcap_{i=1}^\infty A_{ij} $

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    Have you tried to prove these?2012-12-12

1 Answers 1

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Yes, they are always true.

Take $x\in \bigcup_{i=1}^{\infty}\bigcup_{j=1}^{\infty}A_{ij}$. Hence there exists $i$ so that $x\in \bigcup_{j=1}^{\infty}A_{ij}$, and furthermore we find $j$ so that $x\in A_{ij}$. So for this $j$, $x\in \bigcup_{i=1}^{\infty} A_{ij}$, and hence $x\in\bigcup_{j=1}^{\infty}\bigcup_{i=1}^{\infty}A_{ij}$. With identical steps you can conclude the other inclusion as well.

For the second, take $x\in \bigcap_{i=1}^{\infty}\bigcap_{j=1}^{\infty}A_{ij}$. Hence for all $i$, $x\in\bigcap_{j=1}^{\infty}A_{ij}$, and thus for all $i$ and for all $j$, $x\in A_{ij}$. So for all $j$, $x\in\bigcap_{i=1}^{\infty}A_{ij}$, and thus $x\in\bigcap_{j=1}^{\infty}\bigcap_{i=1}^{\infty}A_{ij}$. The other inclusion is analogous.

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    On the other hand $\bigcup_{i=1}^\infty\bigcap_{j=1}^\infty A_{ij}$ does _not_ in general equal $\bigcap_{j=1}^\infty\bigcup_{i=1}^\infty A_{ij}$.2012-12-12