Let $f: X \to Y$ and $a \in X$. Show that if $f$ is continuous at $a$, then if $A \subset X$ and $a \in \overline{A} \implies f(a) \in \overline{fA}$.
I was told that this it is enough to show that if $V \subset Y$ is a neighborhood of $f(a)$ it’s enough to show that $V \cap fA$ isn’t empty, but how is this enough? What if $f(a) \in V$, but $f(a) \notin fA$? The intersection could still be non-empty?
A step back: in any space $X$ and any $B \subseteq X$, $x \in \overline{B}$ iff for every neighbourhood $O$ of $x$, we have $O \cap B \neq \emptyset$. (the closure is the so-called set of adherence points of $B$).
This general fact is just applied to $X=Y$, $B = f[A]$ (or $fA$ if you prefer) and $x=f(a)$.
Proof
Consider the function $f$ as follows:
$f:(X,\tau_1)\rightarrow (Y,\tau_2)$, this to specify the topologies of each space
Let $U\in \tau_2$ such that $f(a)\in U$. We want to prove that $U\cap f(A)\neq \emptyset$. Then let us proceed by contradiction, i.e. $U\cap f(A)=\emptyset$
Note the following
$$ U\cap f(A)=\emptyset \Rightarrow U\subset Y-f(A) \Rightarrow f^{-1}(U)\subset f^{-1}(Y-f(A))=X-f^{-1}(f(A)) \text{ }... \text{ } (1) $$ From the basic properties of functions we know that $A\subset f^{-1}(f(A))$ then $X-f^{-1}(f(A)\subset X-A$. Therefore from $(1)$ we have $f^{-1}(U)\subset X-A$ and $f^{-1}(U)\cap A=\emptyset$
Now, as $f$ is continuous in $a$, for the open set $U$ there is $V\in \tau_1$ such that $a\in V$ and $f(V)\subset U$. Again by the basuc properties of functions we have
$$ V\subset f^{-1}(f(V))\subset f^{-1}(U) \Rightarrow V\cap A\subset f^{-1}(U)\cap A=\emptyset \Rightarrow V\cap A=\emptyset $$
We have a contradiction because $a\in \overline{A}$ and for the open set $V$ we have $V\cap A\neq \emptyset$
Finally $f(a)\in \overline{f(A)}$
