TITLE: Some very weak statements on choice
QUESTION [2 upvotes]: This is a follow-up question to Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?
Consider the statements


$(\text{S}1)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\cup\{X\})$ into $X$.


and


$(\text{S}2)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\times\{0\}) \cup (X\times\{1\})$ into $X$.


Is it true that ${\sf ZF} < {\sf ZF}+(\text{S1}) < {\sf ZF}+(\text{S}2)$? (Here $<$ means "strictly weaker".)

REPLY [4 votes]: $S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.
$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.
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