tag:blogger.com,1999:blog-308061802024-03-13T17:27:44.424-04:00Are you bored yet?eteisworthhttp://www.blogger.com/profile/14461729095493476974noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-30806180.post-6656064327751922232019-03-01T11:52:00.002-05:002022-11-18T20:49:53.206-05:00On a result of Gitik and Shelah\(<br />
\DeclareMathOperator{\pp}{pp}<br />
\DeclareMathOperator{\tcf}{tcf}<br />
\DeclareMathOperator{\pcf}{pcf}<br />
\DeclareMathOperator{\cov}{cov}<br />
\def\cf{\rm{cf}}<br />
\DeclareMathOperator{\otp}{otp}<br />
\def\REG{\sf {REG}}<br />
\def\restr{\upharpoonright}<br />
\def\bd{\rm{bd}}<br />
\def\subs{\subseteq}<br />
\def\cof{\rm{cof}}<br />
\def\ran{\rm{ran}}<br />
\DeclareMathOperator{\ch}{Ch}<br />
\DeclareMathOperator{\PP}{pp}<br />
\DeclareMathOperator{\Sk}{Sk}<br />
\)<br />
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Granted, the title of this post is not very illuminating, as there are several theorems due to Gitik and Shelah. What I have in mind is the following result:</div>
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<b>Theorem 1.5 of [GiSh:412]</b></div>
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<i><b>If $\mu>\kappa$, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.</b></i></div>
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The result is in [GiSh:412], and is described as a ``non-GCH version'' of the Hajnal-Shelah theorem stating the set $\{\mu^\theta:2^\theta<\mu\}$ is finite. It is a lovely illustration of the power of Shelah's cov vs. pp Theorem, and it is the use of this theorem that requires the fourth parameter in cov to be uncountable.</div>
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The proof in [GiSh:412] is very short, but it doesn't quite work as written. We'll take a look at their proof and show it works except in the case where $\kappa$ is singular of uncountable cofinality, and then give an argument reducing this case to the one where $\kappa$ is regular. </div>
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<b>Lemma</b></div>
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If $\mu>\kappa$ and $\kappa$ is not singular of uncountable cofinality, then the set $\{\cov(\mu,\kappa,\theta,\omega_1):\theta<\kappa\}$ is finite.</div>
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<i>Proof:</i></div>
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Assume this fails. By monotonicity properties of cov, we can find an increasing sequence $\langle\theta_n:n<\omega\rangle$ of cardinals such that</div>
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$$\cov(\mu,\kappa,\theta_n,\omega_1)<\cov(\mu,\kappa,\theta_{n+1},\omega_1)$$</div>
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for each $n<\omega$. Without loss of generality, we may assume each of these covering numbers is greater than $\mu$.</div>
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Since we have $\omega_1$ appearing as the fourth component, we can apply the Cov vs. pp Theorem (Theorem 5.4 of Chapter II of <i>Cardinal Arithmetic)</i>, which tells us that if $\cov(\mu,\kappa,\theta,\omega_1)>\mu$, then</div>
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$$\cov(\mu,\kappa,\theta,\omega_1)=\sup\{\pp_{\Gamma(\theta,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta\}$$</div>
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In particular, for each $n<\omega$ we have</div>
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$$\mu<\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}.$$</div>
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Given $n$, let $\chi_n$ be the least cardinal satisfying</div>
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<li>$\kappa\leq\chi_n\leq\mu$,</li>
<li>$\omega_1\leq\cf(\chi_n)\leq\theta_n$, and</li>
<li>$\pp_{\Gamma(\theta_n,\omega_1)}(\chi_n)>\mu$.</li>
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Notice that $\chi_n$ must exist as $\cov(\mu,\kappa,\theta_n,\omega_1)>\mu$, and furthermore $\kappa<\chi_n$, because we have assumed $\kappa$ is not singular of uncountable cofinality -- this is the point where we use that assumption.</div>
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Since the sequence $\langle \chi_n:n<\omega\rangle$ is non-increasing, we can drop a few terms and assume that it is constant, say with value $\chi^*$. Thus, there is a single cardinal $\chi^*$ such that for each $n<\omega$, $\chi^*$ is the least cardinal satisfying</div>
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<li>$\kappa<\chi^*\leq\mu$,</li>
<li>$\omega_1\leq \cf\chi^*<\theta_n$, and</li>
<li>$\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)>\mu$.</li>
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Now suppose $\kappa<\chi<\chi^*$ and $\omega_1\leq\cf(\chi)<\theta_n$. Since $\chi^*$ is the least cardinal with the enumerated properties, we know </div>
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$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq\mu.$$</div>
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But we can then conclude</div>
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(*) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\chi^*$,</div>
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as otherwise Inverse Monotonicity (see part (4) of the No Hole Conclusion 2.3 of Chapter II of <i>Cardinal Arithmetic</i>) forces us to conclude</div>
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$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)\leq\pp_{\Gamma(\theta_n,\omega_1)}(\chi),$$</div>
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which would contradict our choice of $\chi^*$.</div>
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Since we have (*) holding for all relevant $\chi<\chi^*$, we can now apply a generalization of one of the main results of [Sh:371]. The actual theorem we need is Lemma 1.2 of the published version of [Sh:410], which tells us that since (*) holds for all sufficiently large $\chi<\chi^*$, we have</div>
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(**) $\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*)=\pp_{\Gamma((\cf \chi^*)^+,\omega_1)}(\chi^*).$</div>
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(There are some issues with the proof of Lemma 1.2 in [Sh:410] that I will deal with in a future post, but the result still holds.)</div>
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Thus, the sequence $\langle\pp_{\Gamma(\theta_n,\omega_1)}(\chi^*):n<\omega\rangle$ is actually constant with value some $\lambda^*$. The following claim immediately yields a contradiction:</div>
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<b>Claim:</b> <i>For all $n<\omega$, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$.</i></div>
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<i>Proof of claim:</i></div>
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Let $n<\omega$ be given. Recall that by the cov vs. pp Theorem, we know </div>
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$$\cov(\mu,\kappa,\theta_n,\omega_1)=\sup\{\pp_{\Gamma(\theta_n,\omega_1)}(\chi): \kappa\leq\chi\leq\mu \text{ and }\cf(\chi)<\theta_n\}$$</div>
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so in particular $$\lambda^*\leq\cov(\mu,\kappa,\theta_n,\omega_1).$$</div>
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Let $\chi$ be one of the cardinals used in computing the supremum. By our assumptions on $\kappa$, we know $\kappa\neq\chi$ so $\kappa<\chi$. If $\chi<\chi^*$, then by (*) we know</div>
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$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)<\lambda^*$$</div>
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On the other hand, if $\chi^*<\chi$ then because $\chi<\mu<\pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)$, we know<br />
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$$\pp_{\Gamma(\theta_n,\omega_1)}(\chi)\leq \pp_{\Gamma(\theta_n, \omega_1)}(\chi^*)=\lambda^*$$<br />
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by inverse monotonicity.<br />
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Thus, $\cov(\mu,\kappa,\theta_n,\omega_1)=\lambda^*$. $_\square$<br />
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Now where did we use the assumption that $\kappa$ is not a singular cardinal of uncountable cofinality? It was used when we needed to know $\kappa<\chi^*$, so that we can conclude (*) holds for all sufficiently large $\chi<\chi^*$ of the right cofinality.<br />
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Now how do we deal with this last case? Let us assume that $\kappa$ is singular of uncountable cofinality, and find the sequence $\langle \theta_n:n<\omega\rangle\rangle$. For each $n<\omega$ there is a cardinal $\kappa_n<\kappa$ such that<br />
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$$\cov(\mu,\kappa,\theta_n, \omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)$$<br />
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for all $\kappa^*\geq\kappa_n$ (as the covering numbers are non-increasing as we increase $\kappa^*$ to $\kappa$).<br />
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Since $\kappa$ has uncountable cofinality, there is a single $\kappa^*<\kappa$ (which can be taken to be regular) such that<br />
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$$\cov(\mu,\kappa,\theta_n,\omega_1)=\cov(\mu,\kappa^*,\theta_n,\omega_1)\text{ for all }n<\omega.$$</div>
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Now we generate a contradiction as in the preceding case. $_\square$<br />
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eteisworthhttp://www.blogger.com/profile/14461729095493476974noreply@blogger.com0