What is the smallest $i$ such that ZFC proves $\mathfrak c\le \aleph_i$?

Question

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?

Answer 1

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.