Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order

The separation axioms are the part of general topology where textbooks disagree most sharply about vocabulary, and where a reader arriving with the other convention misreads statements rather than merely finding them unfamiliar. This remark settles the disagreements that are live on this page, records the one implication of the classical chain that this page does not prove, and states the choice cost of the one implication whose proof spends a choice principle. The standing topological vocabulary is used throughout: neighbourhoods need not be open, empty intersections equal the whole carrier, a basis is always relative to a topology, and comparisons use coarser and finer.

They do not, in this library. Regular, completely regular, normal, completely normal and perfectly normal name separation conditions on sets alone (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly, Completely regular spaces and Tychonoff ($T_{3\frac{1}{2}}$) spaces, Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, Completely normal ($T_5$) and perfectly normal ($T_6$) spaces); the numerals $T_3$, $T_{3\frac{1}{2}}$, $T_4$, $T_5$ and $T_6$ name the conjunction of each with $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

Munkres builds $T_1$ into regular and normal and then has no separate name for the bare conditions; Kelley, Willard and Engelking take the side taken here. Both usages are current, and neither is more correct. The reason for the choice made here is that the two halves are genuinely independent and each is used alone on this page: the indiscrete topology on a two-point set is regular, completely regular, normal, completely normal and perfectly normal, and fails $T_0$; and the cofinite topology on an infinite set is $T_1$ and fails everything above it. Every statement on this page writes the $T_1$ hypothesis out where it is used, so a reader may translate to the other convention by deleting it.

The word Tychonoff is used for completely regular plus $T_1$, and $T_{3\frac{1}{2}}$ is treated as a synonym.

• Urysohn space, $T_{2\frac{1}{2}}$: distinct points have neighbourhoods with disjoint closures (Urysohn ($T_{2\frac{1}{2}}$) space: distinct points have neighbourhoods with disjoint closures). This is what "Urysohn" means on this page.
• Completely Hausdorff: distinct points are separated by a continuous real-valued function. Some texts attach Urysohn's name to this condition instead. This library does not define it.
• Urysohn's lemma: the theorem, proved later under dependent choice, that in a normal $T_1$ space two disjoint closed sets are separated by a continuous function into $[0,1]$. It is a theorem about sets, not points, and it is unrelated to either space condition.

A statement quoting "Urysohn" without saying which is meant is ambiguous; this page always says which.

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 assembles every implication proved here. Against the classical chain it is short by exactly one arrow:

$T_4\Rightarrow T_{3\frac{1}{2}}$: a normal $T_1$ space is completely regular.

This is Urysohn's lemma, applied to the point $\{x_0\}$ — which is closed by $T_1$ — and the closed set $C$. Its proof indexes a family of open sets by the dyadic rationals, choosing each from the previous one by the shrinking lemma; it is not available at this point in the reading order, and no theorem of this page proves it. Where it is named — in Completely regular spaces and Tychonoff ($T_{3\frac{1}{2}}$) spaces and in Every completely regular space is regular, and every Tychonoff space is $T_3$ — it is named as the classical arrow that is missing here, and it is never used as a fact in any proof on this page. What would license it is a page proving Urysohn's lemma, which in this library's plan sits above the present one.

The later proof records its hypothesis. The dyadic construction in Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0,1]$, and conversely such a space is normal ↗ assumes dependent choice and identifies the step that assembles all finite levels. That provides a proved route for the missing arrow, with an explicit sufficient hypothesis; this remark asserts no independence result.

Everything else in the classical chain is here. In particular $T_6\Rightarrow T_5$ is proved (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 proved without any Urysohn function: it needs only normality, the $F_{\sigma }$ presentation of open sets, and the Axiom of Countable Choice recorded in §4 below. A reader who expects that arrow also to be unavailable is thinking of the route through "every closed set is a zero set", which does need Urysohn's lemma; the route taken here does not.

Every proof on this page is a theorem of ZF except 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, which assumes the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) and spends it at one step, selecting one open set for each member of a countable family of closed sets. The hypothesis is written into that theorem's own statement and into clause 1 of 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, and it is inherited by nothing else: in particular the metric results are choice free, so "metrizable implies perfectly normal, completely normal and normal" needs no choice at all, even though the general arrow from perfect to complete normality does.

• Compactness. "A compact Hausdorff space is normal" is the standard first example of a normal space, and it is absent here for a narrower reason than before: general topological compactness itself is now available at this point in the reading order (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), and the two separation lemmas the proof needs are proved there too (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones). What is still absent is the packaged statement itself, which is homed on a page above the present one. What would license restating it here is a home for that packaging above this page, not below it.
• A regular space that is not normal. Every witness reachable from this page's material needs either cardinal arithmetic or the hereditary and productive behaviour of regularity, neither of which is available here. Rather than plant a false statement with no witness, this page omits it; what would license it is a page developing either of those two tools.
• Hereditary and productive behaviour. Which of these axioms pass to subspaces and to products is not asked here. In particular the equivalence "completely normal if and only if hereditarily normal" is not proved, and Completely normal ($T_5$) and perfectly normal ($T_6$) spaces uses only the separated-sets form.
• Zero-set characterisations beyond the metric case. The equivalence "perfectly normal if and only if normal with every closed set a zero set" (Zero sets and cozero sets of continuous real-valued functions) again needs Urysohn's lemma; only the metric direction is proved here, where the distance function supplies the function outright (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).