Can $f(H)$ be closed if $f$ is injective and $H$ is not closed?

Question

Let $f : X \to Y$ be an injective continuous map between topological spaces, and let $H \subseteq X$ be non closed. Can it happen that $f(H)$ is closed?

Answer 1

No, it cannot happen. Consider the set $f^{-1}\bigl(f(H)\bigr)$. It is a closed set, since $f$ is continuous and $f(H)$ is closed. But, since $f$ is injective, $f^{-1}\bigl(f(H)\bigr)=H$.

Answer 2

When writing the question I thought of the answer.

Suppose $f(H)$ is closed but $H$ is not. Then there is a net $x_\alpha \to x \not\in H$ with $(x_\alpha) \subseteq H$. Observe that, as $f$ is injective, $f(x) \not \in f(H)$.

$f$ is continuous, so $f(x_\alpha) \to f(x)$. Moreover, as we have that $\forall \alpha\ f(x_\alpha) \in f(H)$, which is closed, we have that $f(x) \in f(H)$. And that is absurd.