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Are there infinite cardinals $\alpha < \beta$ and a function $f:\beta\to\alpha$ with the following property?

For any $S\subseteq \beta$ with $|S| =\alpha$ the restriction $f|_S:S\to \alpha$ is injective.

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closed as off-topic by Emil Jeřábek, Joseph Van Name, Pace Nielsen, Jan-Christoph Schlage-Puchta, Ben Barber Feb 4 at 12:34

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  • 2
    $\begingroup$ meta.mathoverflow.net/questions/2781/decluttering-mathoverflow $\endgroup$ – Mere Scribe Feb 1 at 10:43
  • $\begingroup$ @Mere: The OP cannot delete a question if it has (1) an answer with an upvote (let alone 4 and a check mark), or (2) two answers. It will not be removed by the Roomba script either. So it really just waits three people to vote to delete it (which may as well include the OP, of course) or for the OP to unaccept the answer, so Joel can delete it, so they can delete the question. $\endgroup$ – Asaf Karagila Feb 5 at 10:57
  • $\begingroup$ First, apologies for the bad question. Then: wouldn't it be disrespectful to unaccept @JDHamkins' answer and ask him to delete it so I can delete this question? I would like to remove it, but I don't want to offend anyone. $\endgroup$ – Dominic van der Zypen Feb 5 at 18:10
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This is impossible. Since $f$ cannot be injective, as $\alpha<\beta$, there will be two points with the same $f$ value, $f(x)=f(y)$, and if we place those two points $x$, $y$ into an $S$ of size $\alpha$, we cannot have $f\upharpoonright S$ injective.

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  • $\begingroup$ Basically, the point is that injectivity is itself a highly local notion, since $f$ is injective if and only if the restriction of $f$ to all two-element subsets of the domain is injective. $\endgroup$ – Joel David Hamkins Jan 31 at 13:26

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