# If $f:X\to Y$ is $1-1$ and continuous on X, then for any subset $S\subset X$, $f(S')\subset f(S)'$

i wonder if this is true. If $f:X\to Y$ is $1-1$ and continuous on X, then for any subset $S\subset X$, $f(S')\subset f(S)'$. If this is not true, any counter example?

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what does $S'$ mean? –  Holdsworth88 Nov 22 '12 at 10:07
@Holdsworth88 derived set of $S$. –  Mathematics Nov 22 '12 at 10:08
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## 1 Answer

(Big) Hint: If $x \in X$ is such that $f(x) \notin ( f[S] )^\prime$, then there is an open neighbourhood $V$ of $f(x)$ such that $V \cap f[S] \subseteq \{ f(x) \} = f [ \{ x \} ]$. Take the inverse image of both sides, and use the fact that $f$ is continuous and injective to find an open neighbourhood $U$ of $x$ such that $U \cap S \subseteq \{ x \}$.

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