$\overline{f^{-1}(E)} \subset f^{-1}(\overline{E})$ iff $f$ is continuous.

Question

If $f: X \rightarrow Y$ is continous, $X,Y$ topological space, then I am looking for an example of $X, Y, f$ such that $f^{-1}(\overline{E}) \not\subset \overline{f^{-1}(E)}$. Does anyone have an example of this?

Thank you

Answer 1

Let $X$ be a topological space with two topologies $\tau$ and $\tau_2$, where $\tau$ is not the discrete topology. If you find $E\subseteq X$ such that $\overline{E}^{\tau}=X$, then $Id_X^{-1}(\overline{E})=X$ but $\overline{E}^{\text{discrete}}=E$. Since every set in the discrete topology is open, $Id_X:(X,\tau_\text{discrete})\to (X,\tau)$, is continuous and satisfies the inequality. For example $E=\mathbb{Q}$ and $X=\mathbb{R}$ with the regular topology.

Answer 2

If $f:X\to Y$ is continuous and $E \subseteq Y$ is a subset, $\overline{E}$ is closed in $Y$ and so by continuity $f^{-1}[\overline{E}]$ is closed in $X$. As $f^{-1}[E] \subseteq f^{-1}[\overline{E}]$ follows from $E \subseteq \overline{E}$ it follows by minimality of closure that $\overline{f^{-1}[E}] \subseteq f^{-1}[\overline{E}]$.

The inclusion can be proper even for continuous and closed $f$: take $f(x)=0$ and $X=Y=\Bbb R$ and $E = (0,\infty)$. Then $f^{-1}[E] = \emptyset$ so $\overline{f^{-1}[E}] = \emptyset$ while $\overline{E} = [0,+\infty)$ and $f^{-1}[\overline{E}] = \Bbb R$.