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If $A,B\subset X$ and $\overline{A}, \overline{B}$ are completely seprated, so also are $A,B$. since $A\subset \overline{A}$, $B\subset \overline{B}$ then, $f(A)\subset f(\overline{A})=0$ and $f(B)\subset f(\overline{B})=1$ for some continuous function $f:X\to [0,1]$. but the converse is true?

TXC
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2 Answers2

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Assume that $A,B\subseteq X$ are completely separated, i.e. there exists a continuous function $f:X\to [0,1]$ with $f(A)=0$ and $f(B)=1$. Recall that continuity is equivalent with $f(\overline{F})\subseteq \overline{f(F)}$ for all $F\subseteq X$. In other words, continuity of $f$ implies $f(\overline{A})\subseteq \overline{f(A)}=\overline{\{0\}}=\{0\}$ and $f(\overline{B})\subseteq \overline{f(B)}=\overline{\{1\}}=\{1\}$. Hence $\overline{A}$ and $\overline{B}$ are also completely separated.

T. Eskin
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HINT: Suppose that $A$ and $B$ are completely separated, and let $f:X\to[0,1]$ be a continuous function such that $f(x)=0$ for all $x\in A$ and $f(x)=1$ for all $x\in B$. Since $f$ is continuous, $f^{-1}[\{0\}]$ is closed, and certainly $A\subseteq f^{-1}[\{0\}]$, so ... ?

Brian M. Scott
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    So $\overline{A}\subseteq \overline{f^{-1}[{0}]}\subseteq f^{-1}[{0}]$ hence, $f(\overline{A})\subseteq ff^{-1}[{0}]\subseteq [{0}]={0}$. – TXC Feb 05 '13 at 07:41
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    @TXC: Exactly. Or even just ‘$f^{-1}[{0}]$ is a closed set containing $A$, so it contains $\operatorname{cl}A$, which $f$ therefore sends to $0$.’ – Brian M. Scott Feb 05 '13 at 08:05