Hereditary and Productive Behaviour of the Separation Axioms: Examples and Counterexamples

The discrete topology is $T_1$, Hausdorff, regular, normal, completely regular, and completely normal: singleton complements and singleton sets give all required separating opens, and every function to $[0,1]$ is continuous. Its subspaces and products keep the positive properties listed in $T_0$, $T_1$, $T_2$, regularity, $T_3$, complete regularity, and Tychonoffness are hereditary and $T_0$, $T_1$, $T_2$, regularity, $T_3$, complete regularity, and Tychonoffness are productive.

For an infinite set with the cofinite topology of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, every singleton is closed, so the space is $T_1$; two nonempty open sets intersect, so it is not Hausdorff. This displays why the positive table does not infer a higher axiom from $T_1$ alone. An ordinal with the order topology of The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis supplies the contrasting clopen order intervals used by the deleted-plank construction.

In a compact Hausdorff space, normality follows from A compact Hausdorff space is regular and normal, hence $T_3$ and $T_4$. Any closed interval inside a compact ordered interval, any closed endpoint set, and every finite closed subset therefore inherit normality through Every closed subspace of a normal space is normal.

The qualification “closed” is essential. The deleted Tychonoff plank on the A page is an open subspace of a compact Hausdorff normal space but is not normal, so the example cannot be strengthened by replacing “closed” with “arbitrary”.

Assume countable choice. In $P=(\omega_1+1)\times(\omega+1)$ remove the corner $(\omega_1,\omega)$. The resulting plank is regular and $T_1$, hence $T_3$, but it contains the closed edge sets $\{\omega_1\}\times\omega$ and $\omega_1\times\{\omega\}$ that have no disjoint open neighbourhoods.

The parent is compact Hausdorff and normal. The obstruction is concrete: a supposed separation gives an ordinal bound for the countably indexed vertical-tail neighbourhoods, and a horizontal-tail neighbourhood beyond that bound meets a vertical-tail neighbourhood. All stated properties, including the location of countable choice, are proved in Assuming countable choice, the deleted Tychonoff plank is a regular nonnormal open subspace of a compact Hausdorff normal space.

Assume choice. The half-open intervals of the lower-limit line are clopen, and the line is normal by Assuming countable choice, the lower-limit line is normal. Its square remains regular by Arbitrary products of regular spaces are regular, yet the antidiagonal $\{(x,-x):x\in\mathbb R\}$ is closed discrete while $\mathbb Q^2$ is dense and countable by The lower-limit plane has a countable dense set and a closed discrete antidiagonal of size $|\mathbb{R}|$.

Jones's bound turns that profile into nonnormality of the square, as proved in Assuming choice, normality is not productive: the normal lower-limit line has a nonnormal square. This separates the productive behaviour of regularity from the nonproductive behaviour of normality without conflating the two separation conditions.

Assuming countable choice, every open subspace of a normal space is normal.

Assuming choice, the product of any finite family of normal spaces is normal.

Every regular space, and every normal space, is $T_1$.