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Per Löwenheim–Skolem theorems, is it a result that we can have a model of $\sf ZFC$ with any increasing cardinality function over its ranks?

Formally, is the following scheme consistent with $\sf ZFC$?

$ f: {\sf Ord \to Card}\\ \forall \alpha \forall \beta: \beta \geq \alpha \to f(\beta) \geq f(\alpha)\\ \implies\\ \exists M: (M \models {\sf ZFC}) \land \forall \alpha \geq \omega: |V_\alpha^M| = f(\alpha) $

if so, would the same result hold for transitive models of $\sf ZFC$?

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  • $\begingroup$ Clearly not. If $f(\alpha+1)>2^{f(\alpha)}$, then this cannot hold. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 21, 2022 at 13:07
  • $\begingroup$ Why? I thought the upward L-K can pump up the cardinality of the next stage to whatever we wish to $\endgroup$
    – Zuhair Al-Johar
    Commented Sep 21, 2022 at 14:31
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    $\begingroup$ In what sense is $f$ inside $M$, then? $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 21, 2022 at 15:57
  • $\begingroup$ @AsafKaragila, $f$ is not inside $M$. $\endgroup$
    – Zuhair Al-Johar
    Commented Sep 21, 2022 at 19:27
  • $\begingroup$ Then my first comment holds. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 21, 2022 at 19:45

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