Examples for subspace of a normal space which is not normal

Question

Are there any simple examples of subspaces of a normal space which are not normal?

I know closed subspace of a normal space is normal, but open subspace in most cases which I can think of are also normal.

Answer 1

One well-known example is $X=(\omega_1+1)\times(\omega+1)$, where $\omega_1$ is the first uncountable ordinal with the order topology, and $\omega$ is the first infinite ordinal. $X$ is the product of two compact Hausdorff spaces, so $X$ is compact and Hausdorff and therefore normal, but $X\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. $X$ is often called the Tikhonov plank, though the name is also sometimes applied to $X\setminus\{\langle\omega_1,\omega\rangle\}$.

Added: We can also appeal to the theorem that a space $X$ is Tikhonov (i.e., $T_1$ and completely regular) iff it is homeomorphic to a subspace of some product of closed unit intervals. Every product of closed unit intervals is compact and Hausdorff, hence normal, but there are many examples of Tikhonov spaces that are not normal, and every non-normal Tikhonov space is an example of a non-normal subspace of a normal space. A couple of the better-known examples of non-normal Tikhonov spaces are the Moore plane and the Sorgenfrey plane. Another example is the space described in this answer, if $\mathscr{D}$ is taken to have cardinality $2^\omega=\mathfrak c$, which the addendum to the answer shows to be possible; non-normality of the space follows from Jones’s lemma.

Answer 2

${\mathbb R}$ is homeomorphic to $(0,1)$ and Tikhonov theorem tells us that the product of countably infinitely many copies of an interval is compact Hausdorff so a subspace of a normal subspace is not always normal.

Answer 3

The topology $T = \{\emptyset, X , \{a\} , \{a ,b\}, \{a , c\}, \{a ,b ,d\}\}$ on $X = \{a, b , c , d \}$ is normal. Take $Y = \{a, b, c\}$. Then the topology on the set Y is $S = \{ \emptyset , Y , \{a\} , \{ a ,b \} , \{ a , c \}\}$ is not normal. Hence subspace of normal space need not be normal.