$T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces

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).

• $X$ is $T_0$, or a Kolmogorov space, when any two distinct points are topologically distinguishable: for all $x, y \in X$ with $x \ne y$ there is an open set containing exactly one of $x$ and $y$.
• $X$ is $T_1$, or a Frechet space, when each of any two distinct points has an open set containing it and missing the other: for all $x, y \in X$ with $x \ne y$ there are $U, V \in \mathcal{T}$ with

$x \in U, \quad y \notin U, \qquad y \in V, \quad x \notin V .$

Nothing is asserted about a pair of equal points, so a space with at most one point satisfies both conditions vacuously.

Since an open set containing a point is an open neighbourhood of it (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open), both conditions may be read with "open neighbourhood" in place of "open set"; and by the same equivalence recorded in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open they may be read with arbitrary neighbourhoods, since a neighbourhood of $x$ contains an open one and an open neighbourhood is a neighbourhood.

Every $T_1$ space is $T_0$, and this is discharged here rather than left to the reader, because it is the bottom arrow of the whole hierarchy on this page. Let $x \ne y$ and take $U, V$ as in the $T_1$ condition. Then $U$ is an open set containing $x$ and not $y$, so it contains exactly one of the two points, which is the $T_0$ condition. Only the first half of the $T_1$ condition is used, so the implication does not reverse formally, and it does not reverse in fact: Sierpinski space is a witness, recorded on the companion page.

The two conditions differ exactly in symmetry. $T_0$ asks for one open set that tells the pair apart, with no control over which of the two it contains; $T_1$ asks for both separations at once. In Sierpinski space $(\{a,b\}, \{\varnothing, \{b\}, \{a,b\}\})$ of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies the open set $\{b\}$ contains $b$ and not $a$, so the space is $T_0$; but the only open set containing $a$ is the whole space, which also contains $b$, so it is not $T_1$.

Neither condition is a property of a set alone. Both are properties of the pair $(X, \mathcal{T})$, and both are inherited upwards along the comparison order of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison: if $\mathcal{T}_1 \subseteq \mathcal{T}_2$ and $(X,\mathcal{T}_1)$ is $T_0$, respectively $T_1$, then so is $(X,\mathcal{T}_2)$, since the separating open sets of the coarser topology lie in the finer one. In particular the discrete topology satisfies both, and the indiscrete topology on a set with at least two points satisfies neither.

• The names. The numbering $T_0, T_1, T_2, \dots$ is Alexandroff and Hopf's Trennungsaxiome; the individual names honour Kolmogorov and Frechet. This page fixes each axiom by its condition and treats the numeral as an abbreviation, because the numerals above $T_3$ are used inconsistently in the literature (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

• What $T_1$ says about closures. $T_1$ is equivalent to the closedness of every singleton, and hence to $\overline{\{x\}} = \{x\}$ for every point (Interior, closure, boundary, exterior, derived set and isolated point in a topological space); that equivalence is the next item and is a theorem, not a restatement. The corresponding characterisation of $T_0$, that distinct points have distinct closures, is not needed on this page and is not proved here.

• No separation is built into the word space. Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison assumes none; every separation property on this page is a hypothesis written out where it is used.