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.
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.
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 $U \cap \overline{A} = \emptyset$.)