Let "F" be closed set. Let us prove that "\\forall A\\subseteq X" we have that "\\overline{F\\cap A}\\subseteq F\\cap \\overline{A}".
Let "x\\in\\overline{F\\cap A}" . Then "x" is a cluster point of the set "F\\cap A". This means that each neighbourhood of "x"
contains some element "y\\in F\\cap A". Therefore, "y\\in F" and "y\\in A", and each neighbourhood of "x" contains an element "y\\in F" and each neighbourhood of "x" contains an element "y\\in A"
Consequently, "x\\in \\overline{F}" and "x\\in \\overline{A}". Since "F" is a closed set, "\\overline{F}=F". We conclude that "x\\in F", and thus "x\\in F\\cap\\overline{A}".
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