# Using Dushnik-Miller to prove that every sigma-compact group is ccc

(This is the write-up for a talk I gave in the Toronto Student Set Theory and Topology seminar on May 2, 2013.)

A couple of weeks ago I gave a talk for the set-theoretic topology course I was in, on the topic of cardinal invariants of topological groups. While I was preparing that presentation I discovered the following fact:

#### Theorem [Tkachenko, 1983] Every $\sigma$-compact group is ccc.

I will present a proof that I have adapted from Tkachenko’s original paper (“Souslin property in free topological groups on bicompacta”) and the proof that appears in Arhangel’skii & Tkachenko’s big purple book (Section 5.3 of Topological Groups and Related Structures). Both proofs involve first proving a Ramsey result about covers of a space, then using this to prove that a particular space has “weak-precalibre $\aleph_1$” (i.e. Property K) which is a property that implies ccc. Learning this proof has been part of my ongoing attempt to learn how Ramsey results show up in topology.

## Things I learned

Reading Tkachenko’s original paper presented me with some novel difficulties, the first of which was understanding what the heck the title was saying.

• What is the “Souslin Property”?
• What is a bicompact space?
• What does $A \cap B = \Lambda$ mean?

Unsurprisingly, the Souslin Property is just the countable chain condition (ccc). A bicompact space is one where every open cover has a finite subcover, that is … the space is compact. From what I can figure out, back in the old days of topology (40s?) a bicompact space was meant to denote a Hausdorff compact space. Finally, $A \cap B = \Lambda$ just means that the two sets are disjoint; I think that it was common in Russian papers to use $\Lambda$ instead of $\emptyset$.

Working out this proof also reminded me that a $\sigma$-compact group is not always a $\sigma$-(compact group) group. 😉 That is, a group that is the countable union of compact sets need not be the countable union of compact groups. The reals with usual addition is a great example. What this means is that to prove Tkachenko’s theorem it is not enough to show that compact groups are ccc and use the fact that a countable union of ccc spaces is again ccc.

## The Proof Strategy

The proof will be broken into two distinct parts. We will first show a Ramsey result about covers which will follow from the Dushnik-Miller Theorem. This will have nothing to do with groups, and will only barely use $\sigma$-compactness.

The second part will involve showing that any $\sigma$-compact group has weak-precalibre $\aleph_1$ (also known as Property K), which implies that the group is ccc.

## Some Preliminary Facts

Tkachenko’s theorem is interesting because it isn’t true without the group structure. There are two common examples of compact Hausdorff spaces that are not ccc.

Example 1. The ordinal space $\aleph_\omega = \bigcup_{n < \omega} (\aleph_n + 1)$ is a $\sigma$-compact, non-compact space with a large collection of disjoint open sets (e.g. the successor ordinals are all isolated in this space). Similarly, $\aleph_\omega + 1$ is a compact space with a large collection of disjoint open sets.

Example 2. The Stone-Cech remainder $\beta \omega \setminus \omega$ is a compact set that contains $2^{\aleph_0}$ many disjoint open sets (e.g. generate open sets given by an almost disjoint family of cardinality $2^{\aleph_0}$).

Recall the following Ramsey results:

• $\aleph_0 \longrightarrow (\aleph_0)^2_n$ (Ramsey)
• $\aleph_1 \longrightarrow (\aleph_1 , \aleph_0)^2$ (Dushnik-Miller)

Ramsey’s classical result says:

“If the enemy presents you with a complete graph on $\aleph_0$ whose edges they have nefariously coloured using two colours, then you can find an infinte complete subgraph all of whose edges are the same colour”.

The Dushnik-Miller theorem says:

“If the enemy presents you with a complete graph on $\aleph_1$ whose edges they have nefariously coloured using two colours, then you can find an uncountable complete subgraph in the first colour, or you can find an infinite complete subgraph in the second colour”.

You might think that there should be an uncountable complete subgraph in at least one of the two colours, but this just isn’t true. In general, when we intend to use Dushnik-Miller we colour a pair of nodes in the first colour if they “play nicely together” and use the second colour if they “play poorly together”. Generally we will know some fact like “there can’t be infinitely many nodes that all pairwise misbehave”, which will guarantee that we have an uncountable monochromatic complete subgraph.

## The Ramsey Portion

First a quick definition

Definition: Let $x \in X$, a set, and $\Gamma$ a cover of $X$. Then $st(x, \Gamma) := \bigcup\{\, K \in \Gamma : x \in K\,\}$ is the “star of $x$ in $\Gamma$”.

Essentially the star just picks out the elements of the cover that hit $x$.

Fact 1. Let $X$ be a set, and let $\{\,\Gamma_\alpha : \alpha < \omega \,\}$ be a family of covers of $X$ such that uniformly each $\vert \Gamma_\alpha \vert \leq N$. Let $\{\, x_\alpha : \alpha < \omega\,\} \subseteq X$ (which don’t have to have any specified relation to the covers; they are just points.) THEN

1. There are distinct $\alpha, \beta$ such that $st(x_\alpha, \Gamma_\beta) \cap st(x_\beta, \Gamma_\alpha) \neq \emptyset$.
2. There is an infinite $I \subseteq \omega$ such that any distinct $\alpha, \beta \in I$ satisfy [i].

Fact 2. Let $X$ be a $\sigma$-compact space, and let $\{\,\Gamma_\alpha : \alpha < \omega_1 \,\}$ be a family of covers of $X$. Let $\{\, x_\alpha : \alpha < \omega_1 \,\} \subseteq X$ THEN

1. There is an uncountable $J \subseteq \omega_1$ such that any distinct $\alpha, \beta \in J$ satisfy [1.i].

The only thing we will really need to prove is Fact 1.i as the rest are corollaries of this (together with the Ramsey results I mentioned earlier). Let’s check that [1.ii] and [2.i] both hold if we assume [1.i].

[proof of 1.ii] This is a very easy application of Ramsey’s Theorem. Colour $[\omega]^2$ with two colours by colouring a pair colour 1 if they satisfy the conclusion of [1.i] and use colour 2 if they don’t. Ramsey’s theorem gives us an infinite monochromatic set $I$ and Fact 1.i tells us that it can’t be monochromatic in colour 2. So it must be monochromatic in colour 1, which is exactly what we were looking for. [QED]

[proof of 2.i] This is just like the previous proof, except we need to use Dushnik-Miller instead of Ramsey’s Theorem. But first we need to make sure that each of the covers are finite and uniformly have cardinality less than some $N$.

Write $X$ as an increasing union of compact sets $F_n$, for $n < \omega$. One of these $F_n$ must contain an uncountable subset of $\{\, x_\alpha : \alpha < \omega_1 \,\}$, and, by compactness of $F_n$, we can find finite subcovers of the corresponding $\Gamma_\alpha$. We can further refine to an uncountable subset so that each $\Gamma_\alpha$ has cardinality less than some $N$. [QED]

Now it just remains to prove Fact 1.i. This says something about “swapping the indices”. We will use Ramsey’s theorem, although it isn’t obvious what colouring we should use.

[proof] Without loss of generality we may assume that each $\Gamma_\alpha$ has cardinality exactly $N$. We then write each cover as $\{\, A_\alpha^i: i \leq N \,\}$. (Here our obstinate use of $\alpha$ as the index for a natural number pays off: it is clear which cover each $A_\alpha^i$ comes from!)

Now define a colouring $\chi : [\omega]^2 \longrightarrow N^2$ by $\chi (\alpha, \beta) := (k,l)$ iff $x_\alpha \in A_\beta^k$ and $x_\beta \in A_\alpha^l$.

(Okay, so I’m fibbing a bit. Really this isn’t right because as stated this colouring is not well-defined. It could be that $x_\alpha \in A_\beta^i \cap A_\beta^j$. So we can either assume that $\Gamma_\alpha$ is a partition – even though open covers are not usually partitions- or we can add the extra assumption in the colouring that $(k,l)$ is the lexicographically-least such pair. It won’t be an issue, as you will see later.)

Now by Ramsey’s Theorem there is an infinite $I \subseteq \omega$ and a colour – say $(k,l)$– such that any pair from $I$ gets coloured $(k,l)$. Notably,  since $I$ is infinite it contains at least 3 elements, say $\alpha < \beta < \gamma$.

This gives us three pairs, and six bits of information, so let’s record it:

$x_\alpha \in A_\beta^k$,      $x_\beta \in A_\alpha^l$;
$x_\alpha \in A_\gamma^k$,      $x_\gamma \in A_\alpha^l$;
$x_\beta \in A_\gamma^k$,      $x_\gamma \in A_\beta^l$;

This tells us some stuff about intersections; in particular:

$x_\beta \in A_\alpha^l \cap A_\gamma^k$

But we can see that $st(x_\alpha, \Gamma_\gamma) = A_\gamma^k$ and $st(x_\gamma, \Gamma_\alpha) = A_\alpha^l$ (Indices to the rescue again!) and these two sets have non-empty intersection (thanks $x_\beta$!). So the pair $(\alpha, \gamma)$ is the witness we were seeking.[QED]

Chris Eagle pointed out in the seminar that this proof of [1.i] also gives an immediate proof of [1.ii].

## Intermission for Snacks and Definitions

We have now proved the Ramsey results that we will need. Here’s the other stuff we will need to show ccc.

Defintion. A topological space $X$ has “weak-precalibre $\aleph_1$” if whenever $\mathcal{A}$ is an uncountable open family, then there is an uncountable subfamily $\mathcal{B} \subseteq \mathcal{A}$ such that $\forall A,B \in \mathcal{B}$ we have $A \cap B \neq \emptyset$. (This is also called Property K, or the Knaster Property.)

It is clear that weak-precalibre $\aleph_1$ implies ccc. Also, to answer the question you are thinking, “precalibre $\aleph_1$” (without the `weak’ prefix) means that you must replace “any two elements have non-empty intersection” with “any finite collection of elements has non-empty intersection”.

#### Theorem [Shakhmatov, 1986] The statement “Every $\sigma$-compact group is precalibre $\aleph_1$” is independent of ZFC.

This is shown by looking at the free Abelian group $F(S)$ where $S$ is the one-point compactification of $\aleph_1$ with the discrete topology. Under $MA+ \neg CH$ we get that $F(S)$ has precalibre $\aleph_1$ (which is equivalent to ccc under MA). In the other direction, adding a Cohen real implies $\aleph_1$ is not precalibre for $F(S)$.

Ivan Khatchatourian also pointed out to me that the word “Shakhmatov” is related to the words “check mate”, which means “the king is dead!”.

## Tkachenko’s Theorem

#### Theorem[Tkachenko, 1983] If $X$ is a $\sigma$-compact group, then $X$ has weak-precalibre $\aleph_1$, and thus it is ccc.

proof. Let $\{\, O_\alpha : \alpha < \omega_1\,\}$ be a collection of non-empty open sets in $X$. (We need to find an uncountable subcollection of this that pairwise intersects.)

Let $f: X^3 \longrightarrow X$ be defined by $f(a,b,c) := ab^{-1}c$, which is continuous.

Now we need to find an uncountable collection of covers that will let us apply fact [2.i]. If you don’t believe that we can do the next thing, just take it for granted, I will explain it afterwards. (It is basically just continuity.)

For each $\alpha < \omega_1$, pick any $x_\alpha \in O_\alpha$ and find an open cover $\Gamma_\alpha$ (of ‘small’ sets) such that: $\forall U \in \Gamma_\alpha$ if $y,z \in U$ then (1) $f(x_\alpha, y,z) \in O_\alpha$; and (2) $f(y,z, x_\alpha) \in O_\alpha$.

We apply Fact [2.i] which gives us an uncountable $J \subseteq \omega_1$ such that for $\alpha, \beta \in J$ we get $st(x_\alpha, \Gamma_\beta) \cap st(x_\beta, \Gamma_\alpha) \neq \emptyset$.

Claim: For all $\alpha, \beta \in J$ we get $O_\alpha \cap O_\beta \neq \emptyset$.

Take any $x \in st(x_\alpha, \Gamma_\beta) \cap st(x_\beta, \Gamma_\alpha)$.

Sub-Claim: $f(x_\alpha, x, x_\beta) \in O_\alpha \cap O_\beta$.

Unwrapping the two statements about stars gives:

• $\exists U \in \Gamma_\alpha$ such that $x, x_\beta \in U$; and
• $\exists V \in \Gamma_\beta$ such that $x_\alpha, x \in V$

Using the defining property of the covers $\Gamma_\alpha$ and $\Gamma_\beta$ gives:

• $f(x_\alpha, x, x_\beta) \in O_\alpha$; and
• $f(x_\alpha, x, x_\beta) \in O_\beta$

[QED]

Here’s the proof of that weird property:

Pick any $x_\alpha \in O_\alpha$.

• Let $R_\alpha := \{\, (a,b) \in X^2 : f(a,b, x_\alpha) \in O_\alpha \,\}$; and
• let $L_\alpha := \{\, (a,b) \in X^2 : f(x_\alpha, a,b) \in O_\alpha \,\}$.

Note that both of these are open subsets of $X^2$ (because $f$ is continuous), and both sets contain the diagonal of $X^2$. Thus $R_\alpha \cap L_\alpha$ is a non-empty open set.

[PICTURE]

We then let $\Gamma_\alpha := \{\, V_y : y \in X\,\}$ where $V_y := \{\, x \in X : (x,y) \in R_\alpha \cap L_\alpha \,\}$, is a horizontal slice. By definition of $R_\alpha$ and $L_\alpha$, we see that the cover $\Gamma_\alpha$ has the desired property.

## Some Final Notes

You might have noticed that we didn’t use a whole lot about $\sigma$-compactness. In fact the only topological thing we needed was that Fact [2.i] holds. It turns out that there is a wider class of topological spaces that satisfy this fact: the $\Sigma$-Lindelof spaces. I know what you’re thinking.

Definition: A family of subsets $\mathcal{F} \subseteq \mathcal{P}(Y)$ separates $X$ from $Y \setminus X$ if for every $x \in X$ and $y \in Y \setminus X$ there is an $F \in \mathcal{F}$ such that $x \in F$ but $y \notin F$.

Definition: A space $X$ is a $\Sigma$-Lindelof space is there is a countable family $\mathcal{F}$ of closed sets in $\beta(X)$ that separates $X$ from $\beta(X) \setminus X$. (In general, the least size of such a family is called its Nagami number.)

Fact: Compact spaces have finite Nagami number (as $\{X\}$ is the family that separates $X$ from $\beta(X) \setminus X$). Similarly, $\sigma$-compact spaces have countable Nagami number. (So every $\sigma$-compact space is also a $\Sigma$-Lindelof space.)

While this property seems a bit weird, it is preserved under dense subgroups and products (which isn’t true of $\sigma$-compactness), which makes it nice.

## Bibliography

A. Arhangel’skii, M. Tkachenko, “Topological Groups and Related Structures”, Amsterdam : Atlantis Press ; Singapore : World Scientific, c2008

D. B. Shakhmatov, “Precalibers of $\sigma$-compact topological groups”, Mat. Zametki, 39:6 (1986), 859–868

M. G. Tkachenko, “Souslin property in free topological groups on bicompacta”, Mat. Zametki, 34:4 (1983), 601-607

## One thought on “Using Dushnik-Miller to prove that every sigma-compact group is ccc”

1. When I learned that $\Lambda$ means nothing, I also learned why $\mathrm{V}$ means everything. Sadly, I don’t remember what obscure historical artefact I was reading when I made that connection…

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