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#### Julio

##### Member

- Feb 14, 2014

- 71

$\overline{A}=\{x\in X: U\cap A\ne \varnothing, \text{ for all } U\in\tau \text{ and } x\in U\}.$

Hello, if $x\in\overline{A}$, then for all $V$ neighborhood of $x$, we have $V\cap A\ne \varnothing.$ My question is if its neccesary the inclusion of sets for showed.