This is a follow-up question to [Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?](https://mathoverflow.net/questions/230260/does-x-times-0-cup-x-times-1-leq-x-for-x-infinite-imply-s)


Consider the statements



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> 
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> > $(\text{S}1)$ 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".)



Some very weak statements on choice
$S1$ 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](https://math.stackexchange.com/questions/393196/the-relationship-of-frak-mm-m-to-ac), which implies that $\sf DC\_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC\_\omega)$.


