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Show that:

For every open set $U$ in a topological space $X$ and every $A\subset X$ we have $$\overline{U\cap \overline{A}}=\overline{U\cap A}.$$

The simple and new proof is welcome. Thanks for any help.

1 Answers 1

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Clearly $\overline{ U \cap \overline{A} } \supseteq \overline{ U \cap A }$, so we need only show the opposite.

Suppose that $x \in \overline{ U \cap \overline{ A } }$, and let $V$ be any open neighbourhood of $x$. Then $V \cap ( U \cap \overline{ A } ) \neq \emptyset$. As $V \cap U$ is open, it then follows that $( V \cap U ) \cap A \neq \emptyset$ , and so $V \cap ( U \cap A ) \neq \emptyset$. Therefore $x \in \overline{ U \cap A }$.

(The only non-trivial step depends on the following fact, easily proved: If $U$ is an open subset of a topological space $X$ and $A \subseteq X$ is arbitrary, then $U \cap A = \emptyset$ implies $U \cap \overline{A} = \emptyset$.)

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    The non-trivial part phrased differently: Since $V\cap(U\cap\overline A)\ne\emptyset$, the open set $V\cap U$ contains a point of $\overline A$, and thus a point of $A$ ($x\in\overline A$ if and only if every open set containing $x$ contains a point of $A$).2012-08-06
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    @Arthur Thanks for your answer.The fact $U\cap A=\emptyset$ implies $U\cap \overline{A}=\emptyset$ is very interesting.2012-08-06
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    @David very appreciate the fact: $x\in \overline{A}$ iff every open set containing $x$ contains a point of $A$.2012-08-06