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I wonder if the following is true:

If $f:X\to Y$ is a one-to-one and continuous function, then $f(S')\subset f(S)'$ for any $S\subset X$.

(Here $S^\prime$ denotes the derived set of $S$: the family of all limit points of $S$.)

If this is not true, can any counter example?

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up vote 3 down vote accepted

This statement is true. Since $f$ is injective it suffices to show that $S^\prime \subseteq f^{-1} [ (f[S])^\prime ]$.

If $x \in X \setminus f^{-1} [ (f[S])^\prime ]$, then $f(x) \notin ( f[S] )^\prime$, and so there is an open neighbourhood $V$ of $f(x)$ which contains no point of $f[S]$ other than (possibly) $f(x)$ itself. By continuity of $f$, $f^{-1} [ V ]$ is an open neigbourhood of $x$, and since $f$ is one-to-one it follows that $f^{-1} [ V ]$ contains no points of $S$ other than (possibly) $x$. Therefore $x \notin S^\prime$.

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