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.
Best Answer
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.