Is it consistent in $\mathsf{ZF}$ that there is an infinite cardinal $\kappa$, cardinals $\alpha, \beta\in\kappa$ and a function $f:\kappa\to \alpha$ such that for each $x\in\alpha$ there is an injective map $i:f^{-1}(\{x\})\to \beta$?
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asked Nov 10, 2018 at 9:17
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$\begingroup$ I think in ZFC one can write $\kappa=|\Gamma_f|=|\coprod_{x\in \alpha}f^{-1}(\{x\})|\leq\alpha\beta=\mathrm{max}(\alpha,\beta)<\kappa$? Which point fails in ZF? $\endgroup$– QfwfqCommented Nov 10, 2018 at 9:40
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1$\begingroup$ @Qfwfq How do you prove $|\coprod_{x\in \alpha}f^{-1}(\{x\})|\leq\alpha\beta$ in $\mathrm{ZF}$ if you don't have a choice function $x \mapsto i_x \colon f^{-1}(\{x\}) \to \beta$? $\endgroup$– Stefan MeskenCommented Nov 10, 2018 at 11:37
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$\begingroup$ Do $\alpha,\beta,\kappa$ have to be alephs? I guess so, because you are treating them as sets? $\endgroup$– bofCommented Nov 10, 2018 at 11:42
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$\begingroup$ @Qfwfq You're welcome. $\endgroup$– Stefan MeskenCommented Nov 10, 2018 at 17:29
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1 Answer
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Yes, this is consistent.
Consider a (transitive) model of $\mathrm{ZF}$ in which $\omega_1$ has countable cofinality. Fix a strictly increasing, cofinal sequence $(\xi_n \mid n < \omega)$ in $\omega_1$. Consider $$ f \colon \omega_1 \to \omega, x \mapsto \min \{n < \omega \mid x < \xi_n \}. $$ $f^{-1}(\{n\})$ is bounded in $\omega_1$ for all $n \in \omega$. Hence there is an injection $$ i \colon f^{-1}(\{n\}) \to \omega. $$
