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.