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

September 22, 2008

Last time, we learned that—in a world in which much is uncertain—at least we can trust continuous maps of Hausdorff spaces to behave nicely with respect to dense subsets.

Or can we? We showed that if $S\subset X$ is dense, then any continuous map $f: X\to Y$ is determined by its values on $S$, which jives well with our intuition about maps $f :\mathbb{R}\to\mathbb{R}$, for example.

But in $\mathbb{R}$ we can go ever farther. For any open set $U\subset\mathbb{R}$, we can construct a bump function that is nonzero on $U$, but zero outside of $U$. It follows that if $S\in\mathbb{R}$ is not dense, then continuous functions $f:\mathbb{R}\to\mathbb{R}$ are not determined by their values on $S$.

Is this true of all Hausdorff spaces?

The answer is yes, but proving it requires some creativity. The brute force approach does not work here—there is no clear way to create a bump function on some arbitrary space. If you consider yourself a point-set topology guru, I encourage you to try to prove the proposition yourself before reading on.

Proposition: If $X$ is a Hausdorff space and $S\subset X$ is not dense, then there exists some Hausdorff space $Y$ and two distinct continuous functions $f,g: X\to Y$ that agree on $S$.

Proof: Let $K=\overline{S}$, and let $Y$ be two copies of $X$ glued together along $K$. Since there are points not in $K$, the two embeddings $f,g:X\to Y$ are distinct, but agree on $K$ by construction.

I have kept the proof (extremely) short to illustrate the beauty of the idea, but there are a few assumptions that need to be justified. The most serious of these is the assertion that $Y$ is Hausdorff. But a little reading on disjoint unions and quotient spaces will reveal that this is not particularly difficult. (though it is very necessary to know that $K$ is closed)

I do not know if this result is as strong as possible. So I leave the question to you, if you hunger and thirst for more than these two articles have provided.

Question: Does there exist some $T_1$ space $X$ and a non-dense subset $S\subset X$ such that any continuous map $f:X\to Y$, where $Y$ is Hausdorff, is determined by its values on $S$?

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

September 14, 2008

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.

## Using Zorn’s Lemma: Baer’s Criterion

September 10, 2008

Zorn’s Lemma: If $(\mathcal{P}, \leq)$ is a partially ordered set with the property that every totally ordered subset (often called a chain) has an upper bound, then there exists a maximal $M\in \mathcal{P}$. (that is, for $N\in\mathcal{P}$, we have $M\not\leq N$)

This “lemma” is a basic tool for dealing with large collections in a systematic fashion. (if you are interested in the background behind it, see the Wikipedia articles on Zorn’s Lemma and the axiom of choice) Here I’ll demonstrate a fun application in module theory, but there are hundreds of other examples throughout algebra and set theory.

To understand this example, you should be vaguely familiar with modules and cyclic modules.

If $R$ is a commutative ring, it turns out that there exist $R$-modules $Q$ with a fascinating property: If $f: A \to Q$ is an $R$-module homomorphism and $A\subset B$, where $A$ and $B$ are $R$-modules, then there exists an $R$-module homomorphism $F: B\to Q$ extending $f$. Such modules $Q$ are called injective modules.

This is a surprising property, because in general we cannot make functions “bigger”. For example, if $R=\mathbb{Z}$, we have a map $2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ taking $2a\mapsto \overline{a}$, but in any map from $\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ we have $2\mapsto \overline{0}$.

It is not at all obvious at first that injective modules exist, (other than the zero module) let alone what they look like. Nevertheless, there is one very simple example: the $\mathbb{Z}$-module $\mathbb{Q}$. Intuitively, we can imagine dividing by elements of $\mathbb{Z}$ to extend the domain of some function, but even in this case there is some difficulty in proving that $\mathbb{Q}$ is injective. We need, as they say, to “develop some theory”.

Theorem: (Baer’s Criterion) Suppose that $Q$ has the property that if $I$ is any ideal of $R$ and $f:I\to Q$ is an $R$-module homomorphism, there exists an $R$-module homomorphism $F:R\to Q$ extending $f$. Then $Q$ is an injective $R$-module.

This is a powerful theorem because it is much easier to describe the ideals of a ring than it is to describe every possible homomorphism from every possible $R$-module. For $R=\mathbb{Z}$ and $Q=\mathbb{Q}$, we see quite quickly that any map $n\mathbb{Z} \to \mathbb{Q}$ can be extended to $\mathbb{Z}\to \mathbb{Q}$, so this theorem shows that $\mathbb{Q}$ is injective.

But how to prove Baer’s Criterion? It helps to start by considering the simplest cases.

For a basic example, suppose that $Ra \subset Rb$ is an inclusion of cyclic modules, (that is, modules generated by a single element) and $f: Ra\to Q$ is a homomorphism. We will show that we can extend to $b$. Let $I=\{r\in R | rb \in Ra\}$. Then the map $g: I\to Q$ defined by $g(r) = f(rb)$ is a homomorphism, so we can extend to $G:R\to Q$, by the hypothesis. Define $F(rb) = G(r)$. Some basic manipulation of the definitions show that $F$ is a well-defined homomorphism that extends $f$.

Unfortunately, modules are not, in general, cyclic. But if $A\subset A + Ra'$ for some $a'\not\in A$, and we have $f:A\to Q$, we can perform a nearly identical construction. Let $I=\{r\in R | rb \in A\}$, and extend $g(r)=f(rb)$ to $G:R\to Q$. Then we can define $F(a + ra') = f(a) + G(r)$. Again, it is easy to show that $F$ is a well-defined homomorphism that extends $f$. (this is actually an elementary kind of pushout)

This shows that we can always increase the domain of $f$ by adding a new element. In particular, this proves Baer’s criterion for finitely generated modules. But how can we extend to argument to the infinitely generated case? (for example, the $\mathbb{Z}$-module $\mathbb{Q}$)

This is where Zorn’s Lemma comes in. Say that $A\subset B$ and $f: A\to Q$ is a homomorphism. We can form the partially ordered set of extensions of $f$. (where $(C,g)\leq (D,h)$ if $C\subset D$ and $h$ extends $g$) Then if $\{(A_i, f_i) | i\in I\}$ is a totally ordered set, we can form $\cup f_i : \cup A_i \to Q$, so this verifies the chain condition. Hence there exists a maximal $f' : A' \to Q$, by Zorn’s Lemma.

But if we have some $b\in B-A'$, we can extend $f'$ to $A' + Rb$, contradicting the maximality of $f'$. So we must have $A'=B$, and we are done.

Notice that Zorn’s lemma takes the place of a wishful induction. We want to say, “We keep adding elements to the domain until we are done,” but induction alone is not powerful enough to make this rigorous.