On strong measure zero sets

December 6, 2013

I meant to write a longer blog entry on strong measure zero sets (on the real line $\mathbb R$), but it is getting too long, so it may take me more than I expected. For now, let me record here an argument showing the following:

Theorem. If $X$ is a strong measure zero set and $F$ is a closed measure zero set, then $X+F$ has measure zero.

The argument is similar to the one in

Janusz Pawlikowski. A characterization of strong measure zero sets, Israel J. Math., 93 (1), (1996), 171-183. MR1380640 (97f:28003),

where the result is shown for strong measure zero subsets of $\{0,1\}^{\mathbb N}$. This is actually the easy direction of Pawlikowski’s result, showing that this condition actually characterizes strong measure zero sets, that is, if $X+F$ is measure zero for all closed measure zero sets $F$, then $X$ is strong measure zero. (Since this was intended for my analysis course, and I do not see how to prove Pawlikowski’s argument without some appeal to results in measure theory, I am only including here the easy direction.) Pawlikowski’s argument actually generalizes an earlier key result of Galvin, Mycielski, and Solovay, who proved that a set $X$ has strong measure zero iff it can be made disjoint from any given meager set by translation, that is, iff for any $G$ meager there is a real $r$ with $X+r$ disjoint from $G$.

I proceed with the (short) proof after the fold.

Set theory seminar – Marion Scheepers: Coding strategies (IV)

October 12, 2010

For the third talk (and a link to the second one), see here. The fourth talk took place on October 12.

We want to show the following version of Theorem 2:

Theorem. Suppose $\kappa$ is a singular strong limit cardinal of uncountable cofinality. Then the following are equivalent:

1. For each ideal $J$ on $\kappa$, player II has a winning coding strategy in $RG(J)$.
2. $2^\kappa<\kappa^{+\omega}$.

Since $2^\kappa$ has uncountable cofinality, option 2 above is equivalent to saying that the instance of ${\sf wSCH}$ corresponding to $\kappa$ holds.

Before we begin the proof, we need to single out some elementary consequences in cardinal arithmetic of the assumptions on $\kappa$. First of all, since $\kappa$ is singular strong limit, then for any cardinal $\lambda<\kappa$, we have that

$\kappa^\lambda=\left\{\begin{array}{cl}\kappa&\mbox{\ if }\lambda<{\rm cf}(\kappa),\\ 2^\kappa&\mbox{\ otherwise.}\end{array}\right.$

Also, since the cofinality of $\kappa$ is uncountable, we have Hausdoff’s result that if $n<\omega$, then $(\kappa^{+n})^{\aleph_0}=\kappa^{+n}$. I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here.

Proof. $(2.\Rightarrow 1.)$ We use Theorem 1. If option 1. fails, then there is an ideal $J$ on $\kappa$ with ${\rm cf}(\left< J\right>,{\subset})>|J|$.

Note that ${\rm cf}(\left< J\right>,{\subset})\le({\rm cf}(J,{\subset}))^{\aleph_0}$, and $\kappa\le|J|$. Moreover, if $\lambda<\kappa$, then $2^\lambda<{\rm cf}(J,{\subset})$ since, otherwise,

$({\rm cf}(J,{\subset}))^{\aleph_0}\le 2^{\lambda\aleph_0}=2^\lambda<\kappa$.

So ${\rm cf}(J,{\subset})\ge\kappa$ and then, by Hausdorff, in fact ${\rm cf}(J,{\subset})\ge \kappa^{+\omega}$, and option 2. fails.

$(1.\Rightarrow 2.)$ Suppose option 2. fails and let $\lambda=\kappa^{+\omega}$, so $\kappa<\lambda<2^\kappa$ and ${\rm cf}(\lambda)=\omega$. We use $\lambda$ to build an ideal $J$ on $\kappa$ with ${\rm cf}(\left< J\right>,{\subset})>|J|$.

For this, we use that there is a large almost disjoint family of functions from ${\rm cf}(\kappa)$ into $\kappa$. Specifically:

Lemma. If $\kappa$ is singular strong limit, there is a family ${\mathcal F}\subseteq{}^{{\rm cf}(\kappa)}\kappa$ with ${}|{\mathcal F}|=2^\kappa$ and such that for all distinct $f,g\in{\mathcal F}$, we have that ${}|\{\alpha<{\rm cf}(\kappa)\mid f(\alpha)=g(\alpha)|<{\rm cf}(\kappa)$.

In my notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions $f:{\rm cf}(\kappa)\to\kappa$, and then replace them with (appropriate codes for) the branches they determine through the tree $\kappa^{{\rm cf}(\kappa)}$. Distinct branches eventually diverge, and this translates into the corresponding functions being almost disjoint.

Pick a family ${\mathcal F}$ as in the lemma, and let ${\mathcal G}$ be a subfamily of size $\lambda$. Let $S=\bigcup{\mathcal G}\subseteq{\rm cf}(\kappa)\times\kappa$. We proceed to show that $|S|=\kappa$ and use ${\mathcal G}$ to define an ideal $J$ on $S$ as required.

First, obviously $|S|\le\kappa$. Since $\kappa<\lambda=|{\mathcal G}|$ and ${\mathcal G}\subseteq{\mathcal P}(S)$, it follows that ${}|S|\ge\kappa$, or else ${}|{\mathcal P}(S)|<\kappa$, since $\kappa$ is strong limit.

Now define

$J=\{X\subseteq S\mid\exists {\mathcal H}\subseteq{\mathcal G}\,(|{\mathcal H}|<\omega,\bigcup{\mathcal H}\supseteq X)\}.$

Clearly, $J$ is an ideal. We claim that $|J|=\lambda$. First, each singleton $\{f\}$ with $f\in{\mathcal G}$ is in $J$, so ${}|J|\ge\lambda$. Define $\Phi:[{\mathcal G}]^{<\aleph_0}\to J$ by $\Phi({\mathcal H})=\bigcup{\mathcal H})$. Since the functions in ${\mathcal G}$ are almost disjoint, it follows that $\Phi$ is 1-1. Let $G$ be the image of $\Phi$. By construction, $G$ is cofinal in $J$. But then

${}|J|\le|{\mathcal G}|2^{{\rm cf}(\kappa)}=\lambda 2^{{\rm cf}(\kappa)}=\lambda$,

where the first inequality follows from noticing that any $X\in J$ has size at most ${\rm cf}(\kappa)$. It follows that $|J|=\lambda$, as claimed.

Finally, we argue that ${\rm cf}(\left< J\right>,{\subset})>\lambda$, which completes the proof. For this, consider a cofinal ${\mathcal A}\subseteq\left< J\right>$, and a map $f:{\mathcal A}\to[{\mathcal G}]^{\le\aleph_0}$ such that for all $A\in{\mathcal A}$, we have $A\subseteq\bigcup f(A)$.

Since ${\mathcal A}$ is cofinal in $\left< J\right>$, it follows that $f[{\mathcal A}]$ is cofinal in ${}[{\mathcal G}]^{\le\aleph_0}$. But this gives the result, because

${}|{\mathcal A}|\ge{\rm cf}([{\mathcal G}]^{\le \aleph_0},{\subset})={\rm cf}([\lambda]^{\le \aleph_0},{\subset})>\lambda$,

and we are done. $\Box$

580 -Partition calculus (7)

May 6, 2009

Let me begin with a couple of updates.

In the last Corollary of the Appendix to lecture I.5, I indicate that in ${{\sf ZF},}$ we have that

$\displaystyle \aleph(X)<\aleph({\mathcal P}^2(X))$

whenever ${\aleph(X)}$ is not ${\aleph_\alpha}$ for some infinite limit ordinal ${\alpha<\aleph_\alpha.}$ In fact,

$\displaystyle {\mathcal P}(\aleph(X))\preceq{\mathcal P}^2(X)$

holds.

This result is best possible in terms of positive results. In Theorem 11 of the paper by John Hickman listed at the end, it is shown that for any such ${\alpha}$ it is consistent with ${{\sf ZF}}$ that there is an ${X}$ with ${\aleph(X)=\aleph_\alpha}$ for which ${\aleph(X)=\aleph({\mathcal P}^2(X)).}$

I also want to give an update on the topics discussed in lecture III.3.

${\mbox{Erd\H os}}$ and Hajnal asked whether it is possible to have infinite cardinals ${\tau,\lambda,\kappa}$ such that

$\displaystyle \tau\rightarrow[\lambda]^\kappa_{\lambda^\kappa}.$

Galvin and Prikry showed (see Corollaries 16 and 18 of lecture III.3) that any such ${\tau}$ must be larger than ${\lambda^\kappa}$ and that ${\kappa<\lambda.}$

Following a nice suggestion of Grigor Sargsyan, we use arguments as in Theorem 9 from lecture III.5 to show that this partition relation cannot hold.

The key is the following:

Lemma 1 If there are infinite cardinals ${\tau,\lambda,\kappa}$ such that ${\tau\rightarrow[\lambda]^\kappa_{\lambda^\kappa},}$ then for every sufficiently large ${\gamma}$ there is an elementary embedding ${j:M\rightarrow V_\gamma}$ such that ${|M|=\lambda^\kappa,}$ ${{\rm cp}(j)<\lambda,}$ ${j(\lambda)=\lambda,}$ and ${{}^\kappa M\subseteq M.}$

Here is a brief sketch:

Proof: By Corollary 20 from lecture III.3, the given relation is equivalent to ${\tau\rightarrow[\lambda]^\kappa_\lambda.}$ Consider a ${\kappa}$-Skolem function ${F:[V_\gamma]^\kappa\rightarrow V_\gamma}$ so that any ${Y\subset V_\gamma}$ closed under ${F}$ is both closed under ${\kappa}$-sequences and an elementary substructure of ${V_\gamma.}$

Use ${F}$ to define a coloring ${G:[\tau]^\kappa\rightarrow\lambda}$ by setting ${G(x)=F(x)}$ whenever ${F(x)\in\lambda,}$ and ${G(x)=0}$ otherwise. By assumption, there is ${H\in[\tau]^\lambda}$ with ${G''[H]^\kappa\ne\lambda.}$ Note that if ${Y}$ is the closure of ${H}$ under ${F,}$ then ${Y\cap\lambda=G''[H]^\kappa\cap\lambda\ne\lambda.}$ But we can assure that ${|H\cap\lambda|=\lambda,}$ and the result follows by taking as ${M}$ the transitive collapse of ${H.}$ $\Box$

One concludes the proof by noting that it is impossible to have such embeddings. For this, it suffices that ${{}^\omega M\subseteq M}$ and that ${M}$ admits a fixed point past its critical point. One then obtains a contradiction just as in Kunen’s proof that there are no embeddings ${j:V\rightarrow V,}$ see Corollary 9 in lecture III.3.

Similarly, Matthew Foreman has shown that there are no embeddings ${j:M\rightarrow V}$ with ${M}$ closed under ${\omega}$-sequences. The reason is that any such embedding must admit a fixed point past its critical point, as can be argued from the existence of scales. See the paper by Vickers and Welch listed at the end for a proof of this result.

On the other hand, it is still open whether one can have embeddings ${j:M\rightarrow V}$ such that ${M}$ computes cofinality ${\omega}$ correctly.

1. The Baumgartner-Hajnal theorem

In Theorem 2 of lecture III.5 we showed the ${\mbox{Erd\H os}}$-Rado result that

$\displaystyle \kappa\rightarrow_{top}(Stationary,\omega+1)^2$

whenever ${\kappa}$ is regular. It is natural to wonder whether stronger results are possible. We restrict ourselves here to the case ${\kappa=\omega_1.}$ Due to time constraints, we state quite a few results without proof.

580 -Partition calculus (3)

April 6, 2009

1. Infinitary Jónsson algebras

Once again, assume choice throughout. Last lecture, we showed that ${\kappa\not\rightarrow(\kappa)^{\aleph_0}}$ for any ${\kappa.}$ The results below strengthen this fact in several ways.

Definition 1 Let ${x}$ be a set. A function ${f:[x]^{\aleph_0}\rightarrow x}$ is ${\omega}$-Jónsson for ${x}$ iff for all ${y\subseteq x,}$ if ${|y|=|x|,}$ then ${f''[y]^{\aleph_0}=x.}$

Actually, for ${x=\lambda}$ a cardinal, the examples to follow usually satisfy the stronger requirement that ${f''[y]^\omega=\lambda.}$ In the notation from Definition 16 from last lecture, ${\lambda\not\rightarrow[\lambda]^\omega_\lambda.}$

The following result was originally proved in 1966 with a significantly more elaborate argument. The proof below, from 1976, is due to Galvin and Prikry.

Theorem 2 (Erdös-Hajnal) For any infinite ${x,}$ there is an ${\omega}$-Jónsson function for ${x.}$