There is a hierarchy of separation axioms which impose successively stronger conditions on a topological space X, all having to do with the existence of open
neighborhoods separating two fixed points or closed sets in X. In a general topology book they are numbered using the letter T (from the German trennbar, separable, although the concept called separable in English is only peripherally related). All the axioms listed below are inequivalent, although examples that show T3 ≠ T3.5 ≠ T4 are a little hard to construct. (Evandar has thoughtfully cited some below.) From weakest to strongest:

X is T0 if, given two distinct points x, y in X, one of them has a neighborhood not containing the other. A T0-space which is not T1 is N (the natural numbers) with the topology whose open sets are [n, ∞) for every n in N.
X is T1 if, given two distinct points x, y in X, each of them has a neighborhood not containing the other. This is the weakest separation axiom which implies that a one-point set is closed. At least in the classical situation, the Zariski topology on Cn (n-dimensional complex affine space) is a T1 topology which is not T2.
This is the Hausdorff axiom: given two distinct points x, y in X, there are disjoint neighborhoods U of x and V of y. Evandar has supplied an example to show that T2 ≠ T3 in his writeup below.
A space is T3 or regular if, given x in X and a closed set F not containing x, there are disjoint open sets U containing x and V containing F.
A space is T3.5 or completely regular if, given x and F as for T3, not only are x and F separated by open sets, but by a continuous function: there is a continuous map f: X → [0,1] such that f(x) = 0 and f(F) = {1}. Sometimes a T3.5 space is called a Tychonoff space (various spellings), after the guy who proved that a product of compact spaces is compact.
A space is T4 or normal if, given two disjoint closed sets F and G, there are disjoint open sets U containing F and V containing G. Unlike all the previous axioms, this one is not inherited by subspaces of X, and products of normal spaces need not be normal (the Sorgenfrey line is an example of the second phenomenon). Every metrizable space is normal. There is no "T4.5" analogous to T3.5; in any normal space, two disjoint closed sets can be separated by a continuous function. This is the content of Urysohn's lemma which is often called the first nontrivial theorem of general topology.
Some extra comments on separation axioms:

T1 Another way of phrasing the T1 axiom is: for each x in X, {x} is closed. It's not hard to show this is equaivalent to the T1 axiom as given above, and it's often much easier to work with and to visualize.

A simpler example of a space which is T1 but not T2 is the set R of real numbers with the topology where the open sets are the empty set and all subets U of R such that R\U is finite. It's clear that singleton sets are closed, so this is T1 by the comment above, but it's not T2 because any two nonempty open sets must have infinitely many real numbers in common.

T2 This axiom, the Hausdorff axiom, is the minimal condition under which sequences (or nets) converge to at most one point. In spaces that are not T2, very strange things can happen sequence-wise. For example, consider the real numbers R with the non-Hausdorff topology as above, and the sequence (an) given by an=n. If x is a real number, then every open set containing x must contain all but finitely many members of the sequence - that's how the open sets are defined. So in fact, the sequence converges to every single real number, which is pretty crazy. This is one reason why the Hausdorff axiom is usually required for spaces in analytic topology.

A fun example, of a space that's Hausdorff but not T3 is the 'slotted plane' and can be constructed as follows:

Let X be the real plane R2. The basic neighbourhoods of a point x in X are the sets of the form U=(B \ L) u {x}, where B is an open disk centred on x and L is a finite set of straight lines through x.

Then it's reasonably clear that X is Hausdorff, since it's finer than the Euclidean topology on R2. Now, notice that the subspace topology induced on lines is discrete and consequently that a half-line without its endpoint, such as {(x,0) in R2 : x>0}, is closed in X. The endpoint (say, (0,0)) of this half-line is also closed, but can't be separated from the line by open sets. So this topology is not T3

T3 There are spaces that are T3 but not T3.5, but quite honestly, they're not the kind of spaces I'd want to meet in a dark alley. Braver topologists than me can find an example in the exercises at the end of section 18 in Willard's General Topology.

T3.5 An example of a space that's T3.5 but not T4 is the Moore plane, which consists of the closed upper real half-plane {(x,y) in R2 : x=>0}, where the basic open neighbourhoods of a point (x,y) are:

  • the open disks centred on (x,y) if y>0
  • the sets of the form {(x,0)} u B where B is an open disk tangential to the x-axis at (x,0)

It's not hard to show this is completely regular, although I'm not going to do it here. But the space is not normal; let C be the set {(x,0) : x is rational} and D the set {(x,0} : x is irrational}. These sets are both closed in the Moore plane, but can't be separated by open sets (essentially because the space is separable).

T4 A useful theorem is that a space which is compact and Hausdorff is T4 - this is one of the most common sources of normal spaces.

T5 This is the name sometimes given to another separation axiom. A space X is said to be completely normal, or T5 if whenever C and D are sets such that closure(C)nD = closure(D)nC = {} then there are disjoint open sets U and V such that C is included in U and D is included in V. This is clearly stronger than normality - in fact, it is equivalent to the statement "every subspace of X is normal".

A nice example of a space that's normal but not completely normal is the product space IJ where I is the closed unit interval and J is an uncountable indexing set. This is compact and Hausdorff by Tychnoff's theorem, and so normal, but isn't completely normal since the subspace (0,1)J is not normal (tricky to show).

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