The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_1$ gives $T_3$; completely regular gives regular; regular with $T_1$ gives Urysohn, hence Hausdorff, hence $T_1$, hence $T_0$; and metrizable gives every one of them

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). The following implications hold, and each is proved by an earlier item of this page.

1. Perfectly normal implies completely normal, assuming the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)).
2. Completely normal implies normal, and perfectly normal implies normal.
3. Normal together with $T_1$ implies $T_3$, that is regular together with $T_1$.
4. Completely regular implies regular, and Tychonoff implies $T_3$.
5. Regular together with $T_1$ implies Urysohn, which implies Hausdorff, which implies $T_1$, which implies $T_0$.
6. Metrizable implies every property named above: a metrizable space is perfectly normal, completely normal, normal, Tychonoff, completely regular, $T_3$, regular, Urysohn, Hausdorff, $T_1$ and $T_0$, with no choice principle used.

Reading the numbered axioms in order, clauses 1 to 5 give

$T_6 \Rightarrow T_5 \Rightarrow T_4 \Rightarrow T_3 \Rightarrow T_{2\frac12} \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0 ,$

the first arrow under $\mathrm{AC}_\omega$, together with $T_{3\frac12} \Rightarrow T_3$.

This is the whole of the classical chain that this page proves, and it is one arrow short of the classical chain. The implication $T_4 \Rightarrow T_{3\frac12}$ — a normal $T_1$ space is completely regular — is Urysohn's lemma and is not available at this point in the reading order. Its absence is recorded, with what would license it, in this page's conventions remark; it is deliberately not asserted here, and no clause above may be read as giving it.

• Every clause above is an implication and none is an equivalence. This page refutes four of the possible converses among its false statements — $T_1$ does not give Hausdorff, normal does not give Hausdorff, Hausdorff does not give regular, and unique sequential limits do not give Hausdorff — and asserts nothing about the others.

• The $T_1$ hypothesis is where the numerals differ from the adjectives. Regular, completely regular, normal, completely normal and perfectly normal carry no $T_1$ in this library; $T_3$, $T_{3\frac12}$, $T_4$, $T_5$ and $T_6$ are the conjunctions with $T_1$. Clauses 3 and 5 are the two places the conjunction is genuinely needed for the next arrow, and they are what makes the numbered chain descend at all.

• The countable choice in clause 1 is inherited, not introduced. It is spent in the proof of Assuming countable choice, every perfectly normal space is completely normal: separated sets in a normal space whose open sets are all $F_\sigma$ can be separated by disjoint open sets and nowhere else on this page; clause 6 in particular is choice free, since the metric proofs construct their open sets explicitly.