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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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Are you sure about the first part? What if I do something silly like mapping the discrete set $\mathbf Z_{\geq 0} \to \mathbf R$ by sending $0 \mapsto 0$, $n \mapsto 1/n$ otherwise? – Dylan Moreland Feb 6 '12 at 4:03
Or, less silly, $f:\mathbb R\to\mathbb R$ by $x\to x^2$ and $A=\{-1\}\cup[0,2]$? – Henning Makholm Feb 6 '12 at 4:13
The second part is false too, consider the map from $\mathbb R \to \mathbb Z_{\geq 0} $ by sending everything to $0$. – azarel Feb 6 '12 at 4:15
I should probably have mentioned that the question had a "true or false" component -- assuming that the statements were true was indeed quite naive! – Emir Feb 6 '12 at 5:22
There really isn't any harm in telling us everything you know about the problem :) – Dylan Moreland Feb 6 '12 at 16:06
up vote 4 down vote accepted

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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