Some extra comments on separation

axioms:

__T___{1} Another way of phrasing the T_{1} axiom is: for each x in X, {x} is closed. It's not hard to show this is equaivalent to the T_{1} axiom as given above, and it's often much easier to work with and to visualize.

A simpler example of a space which is T_{1} but not T_{2} 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 T_{1} by the comment above, but it's not T_{2} because any two nonempty open sets must have infinitely many real numbers in common.

__T___{2} 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 T_{2}, very strange things can happen sequence-wise. For example, consider the real numbers **R** with the non-Hausdorff topology as above, and the sequence (a_{n}) given by a_{n}=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 T_{3} is the 'slotted plane' and can be constructed as follows:

Let X be the real plane **R**^{2}. 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 **R**^{2}. 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 **R**^{2} : 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 T_{3}

__T___{3} There are spaces that are T_{3} but not T_{3.5}, 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.

__T___{3.5} An example of a space that's T_{3.5} but not T_{4} is the Moore plane, which consists of the closed upper real half-plane {(x,y) in **R**^{2} : 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).

__T___{4} A useful theorem is that a space which is compact and Hausdorff is T_{4} - this is one of the most common sources of normal spaces.

__T___{5} This is the name sometimes given to another separation axiom. A space X is said to be completely normal, or T_{5} 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 **I**^{J} 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).