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No. Because strong limits cardinals exist in $\sf ZFC$. These are the $\beth_\alpha$ for a non-successor $\alpha$.

In $\sf ZF$ injections can be split, so if $A\leq B$ then we have $A\leq^\ast B$ as well. For this reason I prefer to use a slightly modified (but equivalent) definition for $\leq^\ast$: $A\leq^\a …

The consistency strength is just that of ZFC. Start with GCH, and do a symmetric collapse of $\aleph_{\omega_1}$ to be $\aleph_2$. Since everything is $\sigma$-closed, you get DC for free, and the re …

In their book, "Introduction to Cardinal Arithmetic", Holz, Steffens, and Weitz define (on p.71 of the second edition) as follows. Assume that $\kappa$ is an infinite and $\lambda$ is an uncountable …

It is a theorem of Kuratowski that the following holds: There exists a surjection $f\colon A\to\omega$ if and only if there exists an injection $F\colon\omega\to\mathcal P(A)$. As Joel says, the …

Yes. First, let's just agree that $|[\kappa]^{<\kappa}|=\kappa^{<\kappa}$. One direction is immediate, in the other direction note that every function in $\kappa^{<\kappa}$ is an element of $[\kappa\t …

Note that if $\frak m\leq^* n$, then $2^\mathfrak m\leq 2^\mathfrak n$, since given a surjection from $N$ onto $M$, the preimages define an injection between the power sets. So, if $\frak m\leq^* n\le …

First let me answer the first question. Yes. It is possible. You can find the proof in Jech "The Axiom of Choice" in Theorems 11.2 and 11.3. Also you might be interested in reading Pincus' paper, …

In Koenig and Yoshinobu's paper, they prove (Theorem 6.1) that $\sf PFA$ is preserved under $\omega_2$-closed forcings, and this should give the wanted result, as described by Joel. Bernhard König …

Starting from $L(\Bbb R)$, we can take a symmetric extension which preserves $\sf DC$ and adds an $\omega_1$-amorphous set, just somewhere far above $\Theta$. So we cannot prove that (1) or (2) hold f …