TITLE: Give an example of non-normal subspace of a normal space.
QUESTION [7 upvotes]: We know that a closed subspace of normal space is normal.  
My question was: why should other subspaces not work and then I came up with a counterexample.
It is peculiar that any subspace of regular space is regular.

REPLY [4 votes]: I'll just take a look at the difference between regularity and normality.
Consider the following equivalent formulation of regularity:

A topological space $X$ is regular if for each $x \in X$ and every open neighbourhood $U$ of $x$ there is a neighbourhood $V$ of $x$ with $x \in V \subseteq \overline{V} \subseteq U$.

Now, if $X$ is a regular space, and $Y$ a subspace of $X$, taking $x \in Y$ and an open (in $Y$) neighbourhood $U$ of $x$, we know that $U = U_0 \cap Y$ for some open $U_0 \subseteq X$.  But the regularity of $X$ now implies that there is an open $V_0 \subseteq X$ with $x \in V_0 \subseteq \overline{ V_0 } \subseteq U_0$.  Note that $V = V_0 \cap Y$ is an open subset of $Y$, and since $\mathrm{cl}_Y ( V ) = \mathrm{cl}_X ( V ) \cap Y$ it follows that $\mathrm{cl}_Y ( V ) \subseteq \mathrm{cl}_X ( V_0 ) \cap Y \subseteq U_0 \cap Y = U$.
Now let's try something similar for normality, using the following formulation:

A topological space $X$ is normal if for each closed $F \in X$ and every open $U \subseteq X$ with $F \subseteq U$ there is an open $V \subseteq X$  with $F \subseteq V \subseteq \overline{V} \subseteq U$.

So suppose that $X$ is normal, and $Y$ is subspace of $X$.
If $F \subseteq Y$ is closed, and $U \subseteq Y$ is open such that $F \subseteq U$, then there is a closed $F_0 \subseteq X$ and an open $U_0 \subseteq X$ such that $F = F_0 \cap Y$ and $U = U_0 \cap Y$.  Now, if $F_0 \subseteq U_0$ then we could go as above to show that there is an open $V_0 \subseteq X$ such that $$F_0 \subseteq V_0 \subseteq \mathrm{cl}_X ( V_0 ) \subseteq U_0$$ and then taking $V = V_0 \cap Y$, note that $V$ is open in $Y$, and $$F = F_0 \cap Y \subseteq V_0 \cap Y = V; \\
\mathrm{cl}_Y ( V ) = \mathrm{cl}_X (V) \cap Y \subseteq \mathrm{cl}_X ( V_0 ) \cap Y \subseteq U_0 \cap Y = U.$$
So how do we justify $F_0 \subseteq U_0$?  Obviously, since counterexamples exist, we cannot. Putting it slightly differently, it can happen that given any open $U_0 \subseteq X$ such that $U_0 \cap Y = U$ there is a point in $\mathrm{cl}_X(F)$ which is not in $U_0$.

Looking one of the basic examples, we see that this happens.
Let $X$ be a countably infinite set and $Y$ an uncountable set.  Pick $x_* \in X$ and $y_* \in Y$ arbitrarily.  Define topologies on $X$ and $Y$ as follows:

each $x \in X \setminus \{ x_* \}$ is isolated, and the basic open neighbourhoods of $x_*$ are of the form $\{ x_* \} \cup A$ where $A \subseteq X \setminus \{ x_* \}$ is co-finite;
similarly for $Y$: each $y \in Y \setminus \{ y_* \}$ is isolated, and the basic open neighbourhoods of $y_*$ are of the form $\{ y_* \} \cup B$ where $B \subseteq Y \setminus \{ y_* \}$ is co-finite.

Then it can be shown that $X \times Y$ with the product topology is normal.  However the subspace $Z = ( X \times Y ) \setminus \{ \langle x_* , y_* \rangle \}$ is not normal.
To examine this subspace with respect to the above idea, note that $$F = \{ \langle x , y_* \rangle : x \in X \setminus \{ x_* \} \} = ( X \setminus \{ x_* \} ) \times \{ y_* \}$$ is a closed subset of $Z$, and $$U = \{ \langle x , y \rangle : x \in X \setminus \{ x_* \} , y \in Y \} = ( X \setminus \{ x_* \} ) \times Y$$ is open in $Z$, and $F \subseteq U$.  It is easy to show that $\mathrm{cl}_{X \times Y} ( F ) = \{ \langle x , y_* \rangle : x \in X \} = X \times \{ y_* \}$, and in particular contains $\langle x_* , y_* \rangle$.  However, if $V_0 \subseteq X \times Y$ is open and $\langle x_* , y_* \rangle \in V_0$, then it must be that there is a cofinite $B \subseteq Y \setminus \{ y_* \}$ such that $\{ x_* \} \times ( \{ y_* \} \cup B ) \subseteq V_0$.  But this means that $V_0 \cap Z \neq U$!
(Of course, the above does not show that $Z$ is not normal. But it does give an indication that the basic elementary obstruction to showing that arbitrary subspace of a normal space are normal can occur. Note that this elementary obstruction can be phrased in a different manner: there can be disjoint closed subsets of the subspace $Y$ whose $X$-closures are not disjoint.)