I can't find the source of this statement now, but I've been trying to prove that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\overline{A}\setminus\overline{B} \subseteq \overline{A\setminus B}.[/tex]

Now [itex]x\in\overline{A}\setminus\overline{B}[/itex] means x is in every closed superset of A but there exists a closed superset of B that doesn't contain x, whereas [itex]x\in \overline{A\setminus B}[/itex] means x is in every closed set that contains every point of A that's not also in B.

I've tried various manipulations involving equivalent definitions of closure, but have yet to find any obvious way to proceed. Any hints?

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# Difference of closures is subset of closure of difference

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