Pedagogical Remarks

September 28, 2011 at 15:21 | Posted in Uncategorized | Leave a comment

In this post, I want to take a stab at “pedagogically reorganizing” the preceding proof. This post should be considered tentative, as I’m just trying to work out what the best definitions should be.

Definition 1 Let {\mu} be a singular cardinal, with {\mathfrak{A}} as usual. Suppose {N} is an elementary submodel of {\mathfrak{A}}, {H} is a family of functions in {\prod(\mu\cap{\sf Reg})}, and {\kappa<\mu}. We say {H} supports {N} beyond {\kappa} if

  • {H\subseteq N\cap\prod(\mu\cap{\sf Reg})}, and
  • for every {\eta\in N\cap (\kappa,\mu)\cap{\sf Reg}} and {\alpha\in N\cap\eta}, there is an {f\in H} such that {\alpha<f(\eta)}.

If I’ve done things right, the following should be true.

Theorem 2 Let {\mu} be a singular cardinal, and assume {\sigma\leq{\rm cf(\mu)}<\theta\leq\kappa<\mu} with {\sigma} infinite. If {\mathcal{F}\subseteq\prod(\mu\cap {\sf Reg})} has the property that for every {Y\in [\mu]^{<\theta}}, there is an elementary submodel {N} of {\frak{A}} such that

  • {Y\cup \{\kappa,\mu\}\subseteq N},
  • {|N|=\kappa}, and
  • {N} is supported past {\kappa} by a subset of {\mathcal{F}} of cardinality less than {\sigma},

then

\displaystyle  {\rm cov}(\mu,\kappa^+,\theta,\sigma)\leq |\mathcal{F}|. \ \ \ \ \ (1)

Proof: Assume the above. Given a finite subset {F} of {\mathcal{F}}, let

\displaystyle  A_F:=\mu\cap{\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup F), \ \ \ \ \ (2)

and define

\displaystyle  \mathcal{P}:=\{A_F: F\in[\mathcal{F}]^{<\aleph_0}\}. \ \ \ \ \ (3)

Clearly {\mathcal{P}} and {\mathcal{F}} are of the same cardinality (our hypotheses certainly imply {\mathcal{F}} is infinite!). Since {\mathcal{P}\subseteq [\mu]^\kappa}, we will be done provided we show that every {Y\in [\mu]^{<\theta}} can be covered by a union of fewer than {\sigma} sets in {\mathcal{P}}.

Given {Y}, let {N} be an elementary submodel as in the assumptions of the theorem, and let {H} be a subset of {\mathcal{F}} such that

  • {|H|<\sigma}, and
  • {H} supports {N} beyond {\kappa}.

Let us define

\displaystyle  N^+:={\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup H), \ \ \ \ \ (4)

and note that Lemma 1 in the preceding post tells us

\displaystyle  N\cap\mu\subseteq N^+. \ \ \ \ \ (5)

But {N^+\cap\mu} is the union of fewer than {\sigma} sets from {\mathcal{P}}, as

\displaystyle  N^+\cap\mu=\bigcup\{A_F: F\in [H]^{<\aleph_0}\} \ \ \ \ \ (6)

and so we’re done.

\Box

What’s the point? Well, the above emphasizes that we can bound covering numbers by the sizes of “supportive” families of functions in {\prod(\mu\cap{\sf Reg})}. This is “what’s going on” in this section of The Book, and I think the proofs Shelah gives are best viewed as ways of constructing “supportive families”. We’ll see how it goes when I look at the next result in the chapter…

Finishing up Claim 3.3

September 21, 2011 at 16:35 | Posted in Uncategorized | Leave a comment

So why does {H} work? Let us define

\displaystyle  N^+:={\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup H); \ \ \ \ \ (1)

we should show

\displaystyle  N\cap\mu\subseteq N^+. \ \ \ \ \ (2)

It should come as no surprise that we’re going to accomplish this using our prior postings on elementary submodels. In particular, we need to jump back to our posting of May 24, where we proved a version of Shelah’s Claim 3.3A. The notation and hypotheses are slightly different in our current setting, and since last May is long time ago, I’ll just formulate another version of things and then prove it.

Lemma 1 Suppose {\mu} is a singular cardinal and {\mathfrak{A}} is as usual. Suppose we have objects {\kappa}, {N}, and {H} such that

  • {N} is an elementary submodel of {\mathfrak{A}},
  • {\kappa} is a cardinal less than {\mu} and in {N},
  • {H\subseteq N\cap \prod({\sf Reg}\cap(\kappa,\mu))}, and
  • for every {\eta\in N\cap {\sf Reg}\cap(\kappa,\mu)} and {\alpha\in N\cap\eta}, there is an {f\in H} such that {f(\eta)>\alpha}.

Then

\displaystyle  N\cap\mu\subseteq N^+:={\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup H). \ \ \ \ \ (3)

Proof: Suppose this fails, and define

\displaystyle  \gamma(*)=\min(N\cap\mu\setminus N^+), \ \ \ \ \ (4)

and

\displaystyle  \beta(*)=\min(N^+\setminus\gamma(*)). \ \ \ \ \ (5)

It should be clear that {\gamma(*)} is well-defined and {\kappa<\gamma(*)<\mu}. It is also obvious that {\gamma(*)<\beta(*)}. We will now show that {\beta(*)<\mu}.

This follows immediately if {{\rm cf}(\mu)\leq\kappa}, as then {N^+\cap\mu} will be unbounded in {\mu}. If {{\rm cf}(\mu)>\kappa}, then

\displaystyle  {\rm cf}(\mu)\in N\cap{\sf Reg}\cap (\kappa,\mu), \ \ \ \ \ (6)

and so ordinals of the form {f({\rm cf}(\mu))} for {f\in H} are cofinal in {N\cap\mu}. But each of these ordinals is also in {N^+} (as {{\rm cf}(\mu)} is in {N^+} together with all members of {H}), and this tells us that {N^+\cap\mu} must contain an ordinal bigger than {\gamma(*)}.

Thus, no matter what the cofinality of {\mu}, we know

\displaystyle  \kappa<\gamma(*)<\beta(*)<\mu. \ \ \ \ \ (7)

Pausing for a moment, we note that the salient property of {\beta(*)} guarantees

\displaystyle  N^+\cap [\gamma(*),\beta(*))=\emptyset. \ \ \ \ \ (8)

This is quite powerful, given that we know {\beta(*)\in N^+}. For example, this immediately implies that {\beta(*)} cannot be a successor ordinal, for example, as its predecessor would violate the above condition. Similarly, {\beta(*)} must have cofinality greater than {\kappa} — if not, then we can use the fact that {\kappa\subseteq N^+} to derive a contradiction.

We’ll finish by showing that {\kappa<{\rm cf}(\beta(*))} also leads to a contradiction. Given the definition of {N^+}, we know that any element of {N^+\cap\mu} belongs to a subset of {\mu} of cardinality at most {\kappa} that is definable in {\mathfrak{A}} from finitely many members of {H}. In particular, there is a set {A} such that

  • {\beta(*)\in A},
  • {|A|\leq\kappa}, and
  • {A} is definable in {\mathfrak{A}} from finitely many members of {H}.

(We discussed this in more detail back in May.)

The definability of {A} guarantees that it is an element of {N} as well, and therefore {\beta(*)\in N} as it can be defined as the minimal element of {A\setminus\gamma(*)}.

Thus {\eta:={\rm cf}(\beta(*))} is an element of {N\cap N^+}. Let us fix a strictly increasing function {h\in N\cap N^+} mapping {\eta} onto a cofinal subset of {\beta(*)}. Let us define

\displaystyle  \delta(*):=h^{-1}(\{\gamma(*)\}). \ \ \ \ \ (9)

Note that {\delta(*)\in N\cap \eta}. We have assumed {\eta>\kappa}, so this means

\displaystyle  \eta\in N\cap{\sf Reg}\cap (\kappa,\mu). \ \ \ \ \ (10)

Thus our assumptions give us an {f\in H} with {\delta(*)<f(\eta)<\eta}.

But {f(\eta)} is an element of {N^+} as well, hence so is the ordinal {h(f(\eta))}. We’ve now got a problem: since {h} is an increasing function from {\eta} into {\beta(*)}, it follows that

\displaystyle  \gamma^*=h(\delta(*))<h(f(\eta))<\beta(*), \ \ \ \ \ (11)

contradicting the fact that {N^+\cap [\gamma(*),\beta(*))=\emptyset}. \Box

Now that we’ve established the lemma, let’s look back at what we need to verify in order to apply it. Looking at the bullet points in the statement of our lemma, all we need to worry about is the fourth one. Suppose {\eta\in N\cap{\sf Reg}\cap (\kappa,\mu)} and {\alpha\in N\cap\eta}. Choose {k} large enough so that

\displaystyle  \{\eta,\alpha\}\subseteq N_k. \ \ \ \ \ (12)

Looking back at the construction, we then have

\displaystyle  \alpha\leq g_k(\eta)=\sup(N_k\cap\eta) \ \ \ \ \ (13)

and our choice of {H_{k}} guarantees the existence of a function {f\in H_{k}\subseteq H} with

\displaystyle  \alpha\leq g_k(\eta)<f(\eta). \ \ \ \ \ (14)

Thus, the hypotheses of the lemma have been satisfied by the construction from the previous post, and we are finished.

Proof of Claim 3.3 continued…

September 21, 2011 at 15:05 | Posted in Uncategorized | Leave a comment

In the last post I told you what I am going to do, and in this post I want to formulate things as a stand-alone lemma:

Lemma 1 Let {\mu} be a singular cardinal, and suppose

\displaystyle \aleph_0<{\rm cf}(\sigma)\leq\sigma\leq{\rm cf}(\mu)<\theta\leq\kappa<\mu.

Further suppose

\displaystyle  \mathcal{F}\subseteq\prod({\sf Reg}\cap(\kappa,\mu)) \ \ \ \ \ (1)

has the property that whenever {\mathfrak{a}} is a subset of {{\sf Reg}\cap(\kappa,\mu)} of cardinality less than {\theta}, for any {g\in\prod\mathfrak{a}} there is a family {\mathcal{F}_0\subseteq \mathcal{F}} of cardinality {<\sigma} such that

\displaystyle  (\forall\eta\in\mathfrak{a})(\exists f\in \mathcal{F}_0)\bigl[g(\eta)<f(\eta)]. \ \ \ \ \ (2)

Then given {Y\in [\mu]^{<\theta}}, there is a set {H\subseteq\mathcal{F}} of cardinality {<\sigma} such that

\displaystyle  Y\subseteq\mu\cap{\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup H), \ \ \ \ \ (3)

where {\mathfrak{A}=\langle H(\chi),\in, <_\chi\rangle} is “as usual”.

I don’t know if I’ll give the entire proof in this post, but I’ll at least get started.

So let {Y} be a subset of {\mu} of cardinality less than {\theta}. We now define (by recursion on {k<\omega}) objects {N_k} and {H_k} such that

  1. each {N_k} is an elementary submodel of {\mathfrak{A}} of cardinality less than {\theta}, and
  2. each {H_k} is a subset of {\mathcal{F}} of cardinality less than {\sigma}.

The construction starts by setting

\displaystyle  N_0:={\rm Sk}_{\mathfrak{A}}(Y\cup\{\mu\}). \ \ \ \ \ (4)

Now assume that we are given the model {N_k}. We assume that our construction has achieved {|N_k|<\theta} and so

\displaystyle  \mathfrak{a}_k:=N_k\cap{\sf Reg}\cap (\kappa,\mu) \ \ \ \ \ (5)

is of cardinality {<\theta}. We let {g_k} be the characteristic function of {N_k} in {\mathfrak{a}_k}, i.e.,

\displaystyle  g_k\in\prod\mathfrak{a}_k, \ \ \ \ \ (6)

and

\displaystyle  g_k(\eta)=\sup(N_k\cap\eta) \ \ \ \ \ (7)

for each {\eta\in \mathfrak{a}_k}. (Notice that this definition makes sense, as {|N_k|<\theta\leq\kappa<\eta}.)

By our hypothesis, there is a family {H_k\subseteq \mathcal{F}} of cardinality {<\sigma} such that

\displaystyle  (\forall\eta\in\mathfrak{a}_k)(\exists f\in H_k)\left[g_k(\eta)<f(\eta)\right]. \ \ \ \ \ (8)

We now define

\displaystyle  N_{k+1}={\rm Sk}_{\mathfrak{A}}(N_k\cup H_k), \ \ \ \ \ (9)

and the construction continues. (Note that {N_{k+1}} is still of cardinality less than {\theta}, as {H_k} had cardinality less than {\sigma}.)

In the end, let us define

\displaystyle  N:=\bigcup_{k<\omega}N_k, \ \ \ \ \ (10)

and

\displaystyle  H:=\bigcup_{k<\omega}H_k. \ \ \ \ \ (11)

At this point, our assumption that {\sigma} has uncountable cofinality becomes important, as this implies

\displaystyle  |H|<\sigma. \ \ \ \ \ (12)

In our next post, we’ll show that this {H} does the job.

Ignore that last post…

September 21, 2011 at 14:25 | Posted in Uncategorized | Leave a comment

Actually, I think I lied to you in the last post, at least as far as our strategy is concerned. I’ll outline another route to the result that is closer to what Shelah does; it’s actually shorter than what I had in mind originally.

Recalling our situation, {\mu} is a singular cardinal, and {\mathfrak{A}=\langle H(\chi), \in, <_\chi\rangle} is as usual. We have also assumed {\aleph_0<{\rm cf}(\sigma)\leq\sigma\leq{\rm cf}(\mu)<\theta\leq\kappa<\mu}. Let us assume that {\mathcal{F}} is a family of functions as required in the definition of {\lambda_2}, i.e., a collection of functions that are “{\sigma}-cofinal in any subproduct of size {<\theta} constructed from {\prod({\sf Reg}\cap(\kappa,\mu)}” (see previous posts for a more precise definition, but things should be clear from the construction).

Given a finite subset {F} of {\mathcal{F}}, we define

\displaystyle  A_F:=\mu\cap{\rm Sk}_{\mathfrak{A}}(\kappa+1\cup\{\mu\}\cup F), \ \ \ \ \ (1)

and define

\displaystyle  \mathcal{P}=\{A_F:F\in [\mathcal{F}]^{<\aleph_0}\}. \ \ \ \ \ (2)

Note that each {A_F} is a subset of {\mu} of cardinality {\kappa}, so

\displaystyle  \mathcal{P}\subseteq [\mu]^\kappa. \ \ \ \ \ (3)

Also, it should be clear that

\displaystyle  |\mathcal{P}|\leq |[\mathcal{F}]^{<\aleph_0}|=|\mathcal{F}|, \ \ \ \ \ (4)

so we will be finished if we can prove that any {Y\in [\mu]^{<\theta}} can be covered by a union of fewer than {\sigma} sets drawn from {\mathcal{P}}.

How will we do this? Given {Y\in [\mu]^{<\theta}}, we will produce a family {H\subseteq \mathcal{F}} satisfying the following two properties:

\displaystyle  |H|<\sigma, \ \ \ \ \ (5)

and

\displaystyle  Y\subseteq\mu\cap{\rm Sk}_\mathfrak{A}(\kappa+1\cup\{\mu\}\cup H). \ \ \ \ \ (6)

This suffices, because

\displaystyle  \mu\cap{\rm Sk}_\mathfrak{A}(\kappa+1\cup\{\mu\}\cup H)=\bigcup\{A_F: F\in [H]^{<\aleph_0}\}, \ \ \ \ \ (7)

and since {|[H]^{<\aleph_0}|<\sigma}, we will have shown that {Y} can be covered by a union of fewer than {\sigma} sets from {\mathcal{P}}.

Chugging along

September 12, 2011 at 16:35 | Posted in Uncategorized | Leave a comment

Keeping the notation the same as in the last post, we begin our quest to prove {\lambda_1\leq\lambda_2}. The proof is not surprising. Start by letting

\displaystyle  F\subseteq\prod({\sf Reg}\cap(\kappa,\mu)) \ \ \ \ \ (1)

be a family of cardinality {\lambda_2} “covering all small products”. We let {\chi} be a sufficiently large regular cardinal, and let {M} be an elementary submodel of {H(\chi)} containing the relevant parameters, with

\displaystyle  F\subseteq M. \ \ \ \ \ (2)

Our plan is to prove that

\displaystyle  \mathcal{P}:=M\cap [\mu]^\kappa \ \ \ \ \ (3)

is a family satisfying the needed “covering requirements”. Said another way, we will prove that any element of {[\mu]^{<\theta}} can be covered by a union of fewer than {\sigma} sets from {\mathcal{P}}. Since

\displaystyle  |\mathcal{P}|=|M|=\lambda_2, \ \ \ \ \ (4)

this will establish {\lambda_1\leq \lambda_2}.

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