Lemma 5.26.7. Let $f : X \to X$ be a surjective continuous selfmap of a Hausdorff topological space. If $f$ is not $\text{id}_ X$, then there exists a proper closed subset $E \subset X$ such that $X = E \cup f(E)$.

Proof. Pick $p \in X$ with $f(p) \not= p$. Choose disjoint open neighbourhoods $p \in U$, $f(p) \in V$ and set $E = X \setminus U \cap f^{-1}(V)$. Then $p \not\in E$ hence $E$ is a proper closed subset. If $x \in X$, then either $x \in E$, or if not, then $x \in U \cap f^{-1}(V)$. Writing $x = f(y)$ (possible as $f$ is surjective). If $y \in U \cap f^{-1}(V)$ then we would have $x = f(y) \in V$ which is a contradiction with $x \in U$. Hence $y \in E$ and $x \in f(E)$. $\square$

Comment #8766 by Peter Fleischmann on

Last line of proof: I see y \in f^{-1}(U) \subseteq E, but not y\in V\subseteq E.

There are also:

• 13 comment(s) on Section 5.26: Extremally disconnected spaces

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).