# The Unapologetic Mathematician

## Separation Axioms

Now that we have some vocabulary about separation properties down we can talk about properties of spaces as a whole, called the separation axioms.

First off, we say that a space is $T_0$ if every two distinct points can be topologically distinguished. This fails, for example, in the trivial topology on a set $X$ if $X$ has at least two points, because every point has the same collection of neighborhoods — $\mathcal{N}(x)=\{X\}$ for all points $x\in X$. As far as the topology is concerned, all the points are the same. This turns out to be particularly interesting in conjunction with other separation axioms, since we often will have one axiom saying that a property holds for all distinct points, and another saying that the property holds for all topologically distinguishable points. Adding $T_0$ turns the latter version into the former.

Next, we say that a space is $R_0$ if any two topologically distinguishable points are separated. That is, we never have a point $x$ in the closure of the singleton set $\{y\}$ without the point $y$ being in the closure of $\{x\}$. Adding $T_0$ to this condition gives us $T_1$. A $T_1$ space is one in which any two distinct points are not only topologically distinguishable, but separated. In particular, we can see that the singleton set $\{x\}$ is closed, since its closure can’t contain any other points than $x$ itself.

A space is $R_1$ if any two topologically distinguishable points are separated by neighborhoods. If this also holds for any pair of distinct points we say that the space is $T_2$, or “Hausdorff”. This is where most topologists start to feel comfortable, though the topologies that arise in algebraic geometry are usually non-Hausdorff. To a certain extent (well, to me at least) Hausdorff spaces feel a lot more topologically natural and intuitive than non-Hausdorff spaces, and you almost have to try to construct pathological spaces to violate this property. Back in graduate school, some of us adapted the term to apply more generally, as in “That guy Steve is highly non-Hausdorff.”

One interesting and useful property of Hausdorff spaces is that the image of the diagonal map $\Delta:X\rightarrow X\times X$ defined by $\Delta(x)=(x,x)$ is closed. To see this, notice that it means the complement of the image is open. That is, if $(x,y)$ is a pair of points of $X$ with $x\neq y$ then we can find an open neighborhood containing the point $(x,y)$ consisting only of pairs $(z,w)$ with $z\neq w$. In fact, we have a base for the product topology on $X\times X$ consisting of products two open sets in $X$. That is, we can pick our open neighborhood of $(x,y)$ to be the set of all pairs $(z,w)$ with $z\in Z$ and $w\in W$, where $Z$ is an open subset of $X$ containing $x$ and $W$ is an open subset containing $y$. To say that this product doesn’t touch the diagonal means that $Z\cap W=\varnothing$, which is just what it means for $x$ and $y$ to be separated by neighborhoods!

We can strengthen this by asking that any two distinct points are separated by closed neighborhoods. If this holds we say the space is $T_{2\frac{1}{2}}$. There’s no standard name for the weaker version discussing topologically distinguishable points. Stronger still is saying that a space is “completely Hausdorff” or completely $T_2$, which asks that any two distinct points be separated by a function.

A space $X$ is “regular” if given a point $x\in X$ and a closed subset $C\subseteq X$ with $x\notin C$ we can separate $\{x\}$ and $C$ by neighborhoods. This is a bit stronger than being Hausdorff, where we only asked that this hold for two singletons. For regular spaces, we allow one of the two sets we’re separating to be any closed set. If we add on the $T_0$ condition we’re above $T_1$, and so singletons are just special closed sets anyhow, but we’re strictly stronger than regularity now. We call this condition $T_3$.

As for Hausdorff, we say that a space is completely regular if we can actually separate $\{x\}$ and $C$ by a function. If we take a completely regular space and add $T_0$, we say it’s $T_{3\frac{1}{2}}$, or “completely regular Hausdorff”, or “Tychonoff”.

We say a space is “normal” if any two disjoint closed subsets are separated by neighborhoods. In fact, a theorem known as Urysohn’s Lemma tells us that we get for free that they’re separated by a function as well. If we add in $T_1$ (not $T_0$ this time) we say that it is “normal Hausdorff”, or $T_4$.

A space is “completely normal” if any two separated sets are separated by neighborhoods. Adding in $T_1$ we say that the space is “completely normal Hausdorff”, or $T_5$.

Finally, a space is “perfectly normal” if any two disjoint closed sets are precisely separated by a function. Adding $T_1$ makes the space “perfectly normal Hausdorff”, or $T_6$.

The Wikipedia entry here is rather informative, and has a great schematic showing which of the axioms imply which others. Most of these axioms I won’t be using, but it’s good to have them out here in case I need them.

January 11, 2008 - Posted by | Point-Set Topology, Topology

## 7 Comments »

1. Do any of these separation axioms imply uniformity — the condition that makes Cauchy sequences definable? Comment by Michael Brazier | January 13, 2008 | Reply

2. I think you might be hinting at the fact that a space is “uniformizable” — that it is homeomorphic to a uniform space with the uniform topology — if and only if it is completely regular.

Unfortunately the situation is a little hairier for metrizability — being homeomorphic to a metric space with the metric topology. One example of a “metrization theorem” is that a second-countable regular Hausdorff space is metrizable. And of course second-countability implies that the space is separable as well. We can weaken this countability axiom to get non-separable metric spaces, but it’s interesting to note that metrizability requires a countability axiom, while uniformizability only requires a separation axiom. Comment by John Armstrong | January 13, 2008 | Reply

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4. […] subsequence converging to that point. Why not a subnet in general? Because metric spaces must be normal Hausdorff (using metric neighborhoods to separate closed sets) and first-countable! And as long as […]

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