Revisions to $\aleph_\omega$ many subsets of $\aleph_\omega$

broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594

Source Link

edited Jan 21, 2023 at 17:00

2.8k
6
28
38

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

deleted 11 characters in body; deleted 1 characters in body; deleted 2 characters in body

Source Link

edited Jul 31, 2011 at 12:44

saf

71
2

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$$$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

added 15 characters in body

Source Link

edited Jul 30, 2011 at 18:50

saf

71
2

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$$\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject:

added 388 characters in body; added 3 characters in body; added 5 characters in body

Source Link

edited Jul 30, 2011 at 12:23

saf

71
2

Loading

added 1057 characters in body; added 5 characters in body

Source Link

edited Jul 30, 2011 at 12:17

saf

71
2

Loading

Source Link

created Jul 30, 2011 at 1:03

saf

71
2

Loading

Stack Exchange Network

Return to Answer