## What’s The Deal With Hausdorff Spaces? (Part I)

If you have spent more than a few minutes with point-set topology, chances are that you have heard the term “Hausdorff space.” The axioms of a topological space are perhaps a little too general, and so most topological theorems impose additional axioms. Here are some of the basic separation axioms:

A space is called $T_0$ if for any two distinct points $x$ and $y$, there exists an open set $U$ that contains exactly one of $x$ and $y$.

A space is called $T_1$ if for any two distinct points $x$ and $y$, there exists an open set $U$ that contains $x$ but not $y$, and an open set $V$ that contains $y$ but not $x$.

A space is called $T_2$, or Hausdorff, if for any two distinct points $x$ and $y$, there exist disjoint open sets $U$ and $V$ such that $U$ contains $x$ and $V$ contains $y$.

It should be clear that Hausdorff spaces are $T_1$, and $T_1$ spaces are $T_0$. One example of a $T_1$ space that is not Hausdorff is the integers, where $U\subset \mathbb{Z}$ is an open set if $\mathbb{Z}-U$ is finite.

There are many, many other separation axioms, in fact, there are increasingly strong axioms called $T_3$, $T_4$, $T_5$, and $T_6$. So why is the Hausdorff axiom so important?

One possible explanation involves dense sets. If $X$ is a topological space, a set $S\in X$ is called dense if every nonempty open set $U\subset X$ contains a point from $S$. For example, $\mathbb{Q}\subset\mathbb{R}$ is dense. Notice that any continuous function $f: \mathbb{R}\to \mathbb{R}$ is determined entirely by its values on $\mathbb{Q}$. This can be generalized:

Proposition: If $X$ is a topological space, $S\subset X$ is dense, and $Y$ is a Hausdorff space, then any continuous function $f: X\to Y$ is determined by its values on $S$.

Proof: We show that two functions that are equal on $S$ are equal everywhere. Suppose that $f, g: X\to Y$ are continuous and not equal. Then $\exists x\in X$ such that $f(x)\neq g(x)$.

By the Hausdorff condition, we can choose disjoint open sets $U$ and $V$ with $f(x)\in U$ and $g(x)\in V$. Then $f^{-1} (U)\cap f^{-1}(V)$ is a nonempty open set in $X$, so it contains a point $s\in S$. Hence $f(s)\in U$ and $g(s)\in V$, so $f(s)\neq g(s)$, and $f\neq g$ on $S$.

Furthermore, there exist $T_1$ spaces $Y$ for which the above proposition fails. In fact, our above example is one case: the space $X=\mathbb{Z}$ with the cofinite topology.

$X$ is $T_1$, which is not so hard to verify. It is also not hard to verify that if $S\subset \mathbb{Z}$ is infinite, then $S$ is dense.

Also, any bijective function $f: X\to X$ is continuous. But a bijection cannot be determined by its value on all but two points, so if we choose $S$ so that $X-S$ contains at least two points, then continuous functions $X\to X$ are not determined by their value on $S$.

I feel that this example gives some insight into why Hausdorff spaces are studied so frequently, or at least why they are “just nice enough” in many circumstances.

And next week, I’ll post an exciting converse to the proposition.

### One Response to What’s The Deal With Hausdorff Spaces? (Part I)

1. […] The Deal With Hausdorff Spaces? (Part II) Last time, we learned that—in a world in which much is uncertain—at least we can trust continuous […]