A normal $T_1$ space is regular, hence $T_3$, hence Urysohn, Hausdorff, $T_1$ and $T_0$

Let $(X, \mathcal{T})$ be a $T_4$ space, that is a normal $T_1$ space (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces). Then $X$ is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly), hence $T_3$, and therefore also Urysohn (Urysohn ($T_{2\frac{1}{2}}$) space: distinct points have neighbourhoods with disjoint closures), Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), $T_1$ and $T_0$.

The $T_1$ hypothesis is not decoration. Normality alone implies none of the conclusions: the indiscrete topology on a two-point set is normal and not even $T_0$, which is recorded among this page's false statements.

• The proof is one line and the hypothesis does all the work. Normality separates two closed sets; the $T_1$ hypothesis is exactly what turns the point into one of them. This is the pattern of every "$T_{n+1}$ implies $T_n$" argument in the chain, and it is why this library never builds $T_1$ silently into the words regular and normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

• The converse is not proved here and is not asserted. Whether a $T_3$ space must be normal is left open on this page: every witness reachable from this page's material would need machinery it does not have, so no false statement asserting a reversal is planted here (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).