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Working in ZF, is it possible that there is no cardinal number such that $\mathbb{R}$ can inject into? For if there exists a cardinal number $\kappa$ such that $\mathbb{R}$ injects into $\kappa$, then $\mathbb{R}$ can be well-ordered using the ordering of $\kappa$. Please let me know if I got something wrong.

If it is possible that there is no such cardinal number, then it feels really weird to me because the class of all cardinals include arbitrarily large sets, and the assertion says that reals may be greater than all cardinals?

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    $\begingroup$ Talking about cardinals in ZF and insisting they can only be ordinals is missing the point of cardinal numbers. $\endgroup$
    – Asaf Karagila
    Jan 20 at 8:09

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It is true that if $\mathbb{R}$ is non-well-orderable (which is consistent with $\mathsf{ZF}$), then $\mathbb{R}$ does not inject into any cardinal - indeed, no cardinal surjects onto $\mathbb{R}$.

However, that does not mean that $\mathbb{R}$ is "bigger than all cardinals." The converse is also true: there is, provably in $\mathsf{ZF}$, a cardinal which does not inject into $\mathbb{R}$, and indeed onto which $\mathbb{R}$ does not surject.

There is a kind of "best cardinal associated to $\mathbb{R}$" in $\mathsf{ZF}$, namely $\Theta: =\sup\{\alpha\in\mathsf{Ord}: $there is a surjection from $\mathbb{R}$ to $\alpha\}$. And in reasonable choiceless situations, $\Theta$ is indeed quite large. But there's no sense in which $\mathbb{R}$ is "bigger than all cardinals."

Really, the right takeaway is that if choice fails we cannot even compare sizes of sets in a linear way.

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    $\begingroup$ Shame on you for not clarifying that "cardinal" should be used differently :-P $\endgroup$
    – Asaf Karagila
    Jan 20 at 11:12

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