Give an example of non-normal subspace of a normal space.

Question

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.

Answer 1

For a not to weird example of a normal space which has a non-normal subspace, let $(Y,\tau_Y)$ be a topological space which is not normal (for example, the Moore plane) and let $p$ be a point not in $Y$. Let $X=Y\cup\{p\}$ and $\tau=\tau_Y\cup \{X\}$. It is not hard to see that $(X,\tau)$ is a topological space.

Next, let $A$ be a non empty subset of $X$. Since the only open set which contains $p$ is $X$ and $X\cap A=A\neq \emptyset$, we have that $p\in \bar{A}$ where $\bar{A}$ denotes the closure of $A$ in $X$. It follows that every non-empty closed subset of $X$ must contain $p$.

From here, we can conclude that $X$ is a normal space: whenever we have to disjoint closed sets $C$ and $D$, then one must be empty and we can take $\emptyset$ and $X$ as the open sets which separate them. But $Y$ is a subspace of $X$ an the topology that it inherits from $X$ is $\tau_Y$. By assumption, $(Y,\tau_Y)$ is not a normal space.

Answer 2

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 $U_0$.

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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.)

Answer 3

Since $\mathbf{R}$ is a locally compact Hausdorff space, its $\mathbf{1}-$ point compactification $\mathbf{R^{*}}=\mathbf{R} \cup \lbrace \infty \rbrace$ gives a compact Hausdorff space(and hence normal). Let $\mathbf{A}=\lbrace \infty \rbrace \cup \mathbf{Q}$ be a subspace of $\mathbf{R^{*}}$ . $\lbrace \mathbf{\infty} \rbrace$ and $(-\infty,0] \cap \mathbf{Q} $ are disjoint closed sets in $\mathbf{A}$ since $\lbrace \mathbf{\infty}\rbrace=\mathbf{R^\mathsf{c}}$ and $(-\infty,0] \cap \mathbf{Q}=(-\infty,0] \cap ( \mathbf{Q} \cup \lbrace \infty \rbrace ) = (-\infty,0] \cap \mathbf{A} $ are closed in $\mathbf{R^{*}}$. If $\mathbf{A}$ was normal as a subspace then we can find disjoint open sets in $\mathbf{U_{1},U_{2}}$ in $\mathbf{A}$ such that $\lbrace \infty \rbrace \subset \mathbf{U_{1}}$ and $(-\infty,0] \cap \mathbf{Q}\subset \mathbf{U_{2}}$. Then $\mathbf{U_{1}} = \mathbf{G_{1}} \cap \mathbf{A} $ and $\mathbf{U_{2}} = \mathbf{G_{2}} \cap \mathbf{A} $ such that $\mathbf{G_{1}}$ and $\mathbf{G_{2}}$ are open in $\mathbf{R^{*}}$.Let $\mathbf{H_{1}}$ and $\mathbf{H_{2}}$ be the basic open sets contained in $\mathbf{G_{1}}$ and $\mathbf{G_{2}}$ respectively such that $\lbrace \infty \rbrace \subset \mathbf{H_{1}} \cap \mathbf{A} \subset \mathbf{G_{1}} \cap \mathbf{A} $ and $(-\infty,0] \cap \mathbf{Q} \subset \mathbf{H_{2}} \cap \mathbf{A} \subset \mathbf{G_{2}} \cap \mathbf{A}$. Since $\infty \in \mathbf{H_{1}} \Rightarrow \mathbf{H_{1}^\mathsf{c}}$ is compact in $\mathbf{R}$. Also $(-\infty,0] \cap \mathbf{Q}=\lbrace $all non-positive rationals$\rbrace \subset \mathbf{H_{2}} \Rightarrow \mathbf{H_{2}}$ is open in $\mathbf{R}$ or $\mathbf{H_{2}^\mathsf{c}}$ is compact in $\mathbf {R}$.
Since $\mathbf{U_{1}} \cap \mathbf{U_{2}} =\phi \Rightarrow (\mathbf{H_{1}} \cap \mathbf{A}) \bigcap (\mathbf{H_{2}} \cap \mathbf{A}) = \phi \Rightarrow \mathbf{H_{1}} \cap \mathbf{H_{2}} \cap \mathbf{A} = \phi \Rightarrow \mathbf{H_{1}^\mathsf{c}} \cup (\mathbf{H_{2}} \cap \mathbf{A})=\mathbf{R^{*}} \ $. But $\mathbf{H_{1}^\mathsf{c}}$ is compact in $\mathbf{R}$ implies that it is closed and bounded. In particular $\mathbf{H_{1}^\mathsf{c}}=[a,b]$ for some $a,b\in \mathbf{R}$. But [a,b] $\cup (\mathbf{H_{2}} \cap \mathbf{A}) \subset [a,b] \cap \mathbf{A} = [a,b] \cup \mathbf{Q} \cup \lbrace \infty \rbrace \ne \mathbf{R^{*}}$ sincew we can find an irrational in $\mathbf{R}$ such that it is not in [a,b]. I guess we are done.THIS IS PLAIN WRONG>THIS SPACE IS METRIZABLE>