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Suppose $(X,d)$ and (X',d') are metric spaces and f:X\rightarrow X' is continuous.

(a) If $A\subseteq X$ and $x_o$ is an isolated point of $A$, then $f(x_o)$ is an isolated point of $f(A)$.

Attempt: So an isolated point of $A$ means that $\exists r>0$ s.t. $B_r(a)\cap A=\{a\}$. Since if $f$ is continuous an open set V\subset X' means that $f^{-1}(V)$ is open in $X$, can I use this fact somehow to show that the map preserves the isolated point?

(a) If $A\subseteq X$, $x_o\in A$ and $f(x_o)$ is an isolated point of $f(A)$ then $x_o$ is an isolated point of $A$.

Attempt: Same deal as above -- I'm not sure if I'm thinking of the right theorem in proving this.

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    There really isn't any harm in telling us everything you know about the problem :)2012-02-06

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Both parts are often false. For the first part, see the comment of Dylan Moreland, which even gives a counterexample that is injective.

For the second part, consider constant functions. All you can say is that if $f(x_0)$ is isolated in $f(A)$, then there is a relatively open subset of $A$ containing $x_0$ where $f$ is constant.

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    @Christos: The function is given in that comment, and it is continuous because the domain is discrete.2018-11-16