I think we have that $\aleph_{\mathfrak c}\ge\mathfrak c$. But are there tighter upper bounds for $\mathfrak c$ in ZFC or no such bound is dependent on ZFC?
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4$\begingroup$ Wikipedia says: "A result of Solovay, proved shortly after Cohen's result on the independence of the continuum hypothesis, shows that in any model of ZFC, if $\kappa$ is a cardinal of uncountable cofinality, then there is a forcing extension in which $2^{\aleph _{0}}=\kappa$." This seems to exclude an explicitly described (in a reasonable way) ordinal $\alpha$ such that $\aleph_\alpha$ is provably larger than $\mathfrak c$. $\endgroup$– hmakholm left over MonicaCommented Sep 21, 2017 at 10:50
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That depends on what you mean by $i$.
Indeed, by the work of Cohen, extended by Solovay and ultimately Easton, we know that there is very little to be said in $\sf ZFC$ on the continuum function. In particular, given a model of set theory $V$, for every ordinal $\alpha$, there is an extension of $V$ with the same ordinals and same $\aleph$s, in which $\frak c\geq\aleph_\alpha$. Therefore in that sense, there is no provable bound on the continuum.
But at the same time, what you write is true. $\frak c\leq\aleph_{\frak c}$. Taking the least cardinal that has $\frak c$ cardinals below it, then by definition this cardinal is at least $\frak c$. Even more is possible, it is possible that $\frak c=\aleph_{\frak c}$.
Now why is that not a "provable bound"? Because the value of $\frak c$ is not absolute and itself not computable. The cardinal $\frak c$ is not a fixed ordinal, so putting it in the index of an $\aleph$ is awkward, to the sense that this will be a different cardinal in different models of set theory, even if they have the same cardinals; whereas $\aleph_1$ is always the same in models which agree about cardinals.
So no, there is no provable bound; and to the extent possible, $\aleph_{\frak c}$ is the best we can do, since we can get equality, so certainly nothing smaller is provably smaller.
If you allow to use $\beth$ numbers, then $\beth_1$ is by definition $\frak c$, but of course, the function mapping each $\beth$ to its $\aleph$ value is not determined by the axioms of $\sf ZFC$. So we're back to where we started.
answered Sep 21, 2017 at 11:18
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$\begingroup$ Is the result in your second paragraph independent of how large $V$ is? If $V$ has standard integers but $\aleph_\alpha\in V$ is larger than the metalevel's $\mathfrak{c}$, then it would seem to produce a contradiction if the model could be extended with enough new reals. I could believe it if $V$ is countable, though. $\endgroup$ Commented Sep 21, 2017 at 11:54
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1$\begingroup$ @Henning: Yes, of course. (This is the downside of forcing, formally speaking it only works "well" over countable transitive models, but in practice, we might as well assume that whenever we have a model of any "reasonable extension of ZFC" (i.e. not ZFC+"ZFC is inconsistent" or something), then the model we care about is a countable transitive model.) $\endgroup$– Asaf Karagila ♦Commented Sep 21, 2017 at 11:56
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$\begingroup$ @AsafKaragila: do you have a reference for the independence of $\aleph_{\mathfrak c} = \mathfrak c$? $\endgroup$ Commented Jan 24, 2019 at 0:22
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2$\begingroup$ @PseudoNeo: Use the theorem of Cohen-Solovay, starting from CH, for any cardinal with uncountable cofinality there is a forcing extension where the continuum is that cardinal. Now take any $\aleph$ fixed point of uncountable cofinality. $\endgroup$– Asaf Karagila ♦Commented Jan 24, 2019 at 1:03