3
$\begingroup$

A scattered space is a space for which every not empty subset has an isolated point (equivalently for $T_1$ spaces, every not empty closed subset has an isolated point).

A compact Hausdorff, not scattered space can be continuously mapped onto $[0,1]$. Conversely if $S$ is a compact Hausdorff, scattered space and $f$ in $C(S,[0,1])$ a surjection, how can a contradiction be derived?

$\endgroup$
1
  • $\begingroup$ Are you talking about metric spaces, or generally speaking about compact Hausdorff spaces? $\endgroup$
    – Asaf Karagila
    Aug 30, 2012 at 13:27

1 Answer 1

5
$\begingroup$

More generally, suppose that $S$ is a compact, scattered, Hausdorff space, and $X$ is a compact Hausdorff space. If $f:S\to X$ is a continuous surjection, then $X$ is scattered.

Proof: Suppose that $A$ is a non-empty subset of $X$ with no isolated points. Let $K=\operatorname{cl}A$; $K$ also has no isolated points. Let $\mathscr{C}$ be a maximal chain in $\{C\subseteq S:f[C]=K\}$, and let $C=\bigcap\mathscr{C}$. Then $f[C]=K$, but if $D$ is a compact proper subset of $C$, then $f[D]\ne K$. $S$ is scattered, so $C$ has an isolated point $p$; let $D=C\setminus\{p\}$. $D$ is compact, so $f[D]\ne K$, and hence $f(p)\notin f[D]$. But since $D$ is compact, $f[D]$ is also compact and therefore closed in $X$, so $f(p)$ is an isolated point of $K$, a contradiction. $\dashv$

In Scattered spaces II Kannan and Rajagopalan show that every topological space is the closed, continuous image of a non-Hausdorff scattered space, that every sequential Hausdorff space is the closed, continuous image of a scattered sequential Hausdorff space, and that $[0,1]$ is the continuous image of a scattered, countably compact, first countable, locally countable Tikhonov space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.