I'm facing this problem: For $f : (X,\mathcal{T}) \to (Y,\mathcal{T}')$ a continuous map between topological space, then for any subset $U \subset X$ is $f(\overline{U}) \subset \overline{f(U)}$.
I know the usual proof, but I´d like is this one is right:
Let be $x \in f(\overline{U})$. So for any open neighborhood $V$ of $x$ is $V \cap f(U) \neq \emptyset$.
Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.
Using $f^{-1}(x) \in \overline{U}$ is $f^{-1}(V) \cap U \neq \emptyset$.
So $f(f^{-1}(V) \cap U) = V \cap f(U) \neq \emptyset$, then $x \in \overline{f(U)}$.
Is this right? I'm not sure.
And here another try:
Suppose exist $x \in f(\overline{U})$ with $x \notin \overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V \cap f(U) = \emptyset$.
For $z \in \overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W \cap A \neq \emptyset$.
Then exist $h \in W \cap A$ and $f(h) \in f(W) \cap f(A)$. So for the adequate $f(W) \subset V$ is $f(W) \cap U \neq \emptyset$ but $V \cap f(U) = \emptyset$. A contradiction.
Is this last one right?
Thanks!
Let $f$ be continuous and suppose $y \in f[\overline{U}]$. So $y=f(x)$ with $x \in \overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] \cap U$ is non-empty (as $x \in \overline{U}$), say that $x' \in f^{-1}[O] \cap U$. But then $f(x') \in f[U]$ and $f(x') \in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y \in \overline{f[U]}$ and the inclusion has been shown.