In the paper The Continuumproblem, Felix Bernstein introduces a new axiom and uses it to conclude the continuum hypothesis.
(1) As the paper is relatively old and the writing style is somehow informal, I am wondering if there is a more exact and concrete proof of the result? (or may please give a better presentation of the argument!)
(2) Is the system introduced by Bernstein consistent? (he does not discuss this in the paper)
This is really a long comment. This paper has been reviewed twice by zbMATH: one by H. B. Curry, which is not informative; another by W. Ackermann, which is in German. The following is the (manipulated) translation of part of his review by DeepL.
... However, a certain sense cannot be connected with this axiom, since it is not said what is to be understood by the existence of an element. If one tries to find this sense in the application of the axiom, one is also disappointed. The point of the proof is that a subset of power $\aleph_1$ is extracted from a certain set of the power of the continuum by multiple application of the axiom of choice, and it is claimed that the selection can be made so arbitrarily that it is impossible to determine whether the subset is a real subset or not without giving any reason for it. (Indeed, in the presence of a proof for $2^{\aleph_0}\neq\aleph_1$ this statement would be possible). In any case, the validity of Cantor's hypothesis remains just as indeterminate after the present remarks as before.
Edit: It seems, in this paper (1905), Bernstein has sketched a proof of the Continuum Hypothesis for a submission to Math. Ann.
Ackermann's above-cited review focuses on the second of Bernstein's proposed rules. However, even the first ("axiom of identity") is broken. Gentling tweaking the language, this axiom reads:
$(\star)\quad$ If $M_1$ is a subset of $M$ and there is no proof in our system that $M_1\not= M$, then $M_1=M$.
(FWIW, with my philosophy hat on the least gentle aspect of this tweaking is my shift from "no proof can exist" to "there is no proof," but I don't think this ultimately affects my point.)
No version of this can possibly work as long as this axiom itself is included in "our system" (which I'll call $\mathsf{B}$). By the diagonal lemma, we can produce formulas defining sets $X_1,X$ such that the system in question proves the appropriately-expressed statement $$X_1=\{y\in X: \mathsf{B}\vdash X_1\not=X\}.$$ Whoops. If we want to avoid diagonalization, just pick some sentence $\theta$ independent of $\mathsf{B}$ and consider $$X=\{1,2\}, X_1=\{x\in X: x=1\vee \theta\}, X_2=\{x\in X: x=1\vee\neg\theta\}.$$ Bernstein's rule $(\star)$ would demand $X_1=X_2=X$, giving a contradiction.
And while considering the "our system" in $(\star)$ as applying only to $\mathsf{B}\setminus(\star)$ fixes this problem, it also prevents Bernstein's desired application.
Stepping back a bit, the whole idea of a "default assumption" present here is much trickier to implement than it may first appear; while obvious now, in the late 30s when Bernstein was writing this wasn't yet fully understood.
