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?

  • $\begingroup$ 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


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.


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