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Let $X$ be normal and $p: X \to X/R$ be a closed and open map. Show that $X/R$ is normal.

First, I don't understand how can it be closed and open at the same time, I'm confused.

I search and a space $X$ is normal if for any $A, B$ closed in $X$ there is an open set $U$ for $A$ and an open set $V$ for $B$ such that $U \cap V = \emptyset$, is that right? So, I have to find two open sets which are disjoint in $X/R$?

Martin Sleziak
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Chan27
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  • It's fine for the map to be a closed map and open map at the same time, just like it's fine for a set to be both open and closed. Don't let the common English definitions of those words throw you off. ;) So the map $p$ in question has the property that if $U$ is open in $X$, then $p(U)$ is open in $X/R$, and if $C$ is closed in $X$, then $p(C)$ is closed in $X/R$. – Mike Pierce Sep 21 '17 at 18:24
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    In Polish the terminology is the same. I like the comparison with the door. In the real life it is either open, or closed. This is not the case for sets. Namely, every topological space contains at least one closed and open set: this is the whole space. I am writing about one, because it could be $X={\varnothing}$. :) – szw1710 Sep 21 '17 at 18:28
  • See also: Image of a normal space under a closed and continuous map is normal. Some of the questions linked there might be of interest, too. – Martin Sleziak Dec 17 '21 at 06:45

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Suppose $f: X \to Y$ is closed and continuous and onto and $X$ is normal. Take $C$ and $D$ disjoint and closed in $Y$, then $f^{-1}[C]$ and $f^{-1}[D]$ are disjoint and closed in $X$ so have disjoint open neighbourhoods $U$ and $V$ so that

$$U \cap V = \emptyset, f^{-1}[C] \subseteq U, f^{-1}[D] \subseteq V$$

Define $O_1 = Y\setminus f[X\setminus U]$, $O_2 = Y\setminus f[X\setminus V]$ which are both open as $f$ is a closed map. ($U$ open so $X\setminus U$ closed, so its image under $f$ closed in $Y$ so its complement in $Y$ open there).

Suppose $y \in O_1 \cap O_2$. Then $y = f(x)$ for some $x$. Then $x \notin X\setminus U$ (or otherwise $y \in f[X\setminus U]$, so $y \notin O_1$, contradiction), hence $x \in U$. By the same reasoning $x \in V$ contradicting $U \cap V = \emptyset$, so no such $y$ can exist and $O_1 \cap O_2$.

Now $C \subseteq O_1$ and $D \subseteq O_2$ are also clear: Suppose e.g. $y \in C$, and let $x \in X$ be such that $f(x) = y$. Then $x \in f^{-1}[C] \subseteq U$, so $x \in U$ and as this holds for all such $x$, $y \notin f[X\setminus U]$, so $y \in O_1$, etc.

Hence $Y$ is normal. We don't need $f$ is open, just continuous and closed.

Henno Brandsma
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