TITLE: Prove that $f(\overline A) \subseteq \overline{f(A)}$
QUESTION [2 upvotes]: Let $f : X \longrightarrow Y$ be continuous, and let $A \subseteq X$. Show that $f(\overline A) \subseteq \overline{f(A)}$.

My attempt
Let $y \in f(\overline A)$.
Then there exists $x \in \overline A$ such that $f(x) = y$.
Now let us take an open neighbourhood $V$ of $y$ in $Y$ arbitrarily.
If we can show that $V \cap f(A) \neq \emptyset$ then our purpose will be served.
Now since $f(x) = y \in V$, we have $x \in f^{-1} (V)$.
Since $x \in \overline A$ and $f^{-1} (V)$ is open in $X$ (since $f$ is continuous), we have that $f^{-1} (V) \cap A \neq \emptyset$.
Let $z \in f^{-1} (V) \cap A$.
Then $f(z) \in V \cap f(A)$, which proves that $V \cap f(A) \neq \emptyset$ and we are done.
Is my reasoning correct at all? Please verify it.

REPLY [2 votes]: Your reasoning is correct.

Alternative route: 
$f$ is continuous and $\overline{f(A)}$ is closed, so we conclude that $f^{-1}\left(\overline{f(A)}\right)$ is closed. 
This with: $$A\subseteq f^{-1}\left(f(A)\right)\subseteq f^{-1}\left(\overline{f(A)}\right)$$allowing us to conclude:$$\overline{A}\subseteq f^{-1}\left(\overline{f(A)}\right)$$
or equivalently:
$$f(\overline{A})\subseteq\overline{f(A)}$$
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