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I am looking for a simple proof (no fine structure, please) of the following:

Let $\lambda$ be a limit ordinal, and $\mu < \lambda$, infinite: If $\mathbf{L}_\lambda \models \texttt{"}\mu \mbox{ is a successor cardinal}\texttt{"}$ then $\mathbf{L}_\mu \models ZF^-$.

Where $\mathbf{L}_\lambda \models \texttt{"}\alpha \mbox{ is a cardinal}\texttt{"}$, means here: there is no surjection $\xi \to \alpha$, in $\mathbf{L}_\lambda$, for $\xi < \alpha$.

And $ZF^-$ is Zermelo-Fraenkel minus Power Set.

The proof I have uses some standard long-winded "condensation" arguments + Admissible sets. So I welcome any ideas.

Short of this, can anyone suggest a short, simple proof of the following:

Let $\lambda$ be a limit ordinal [or even limit of limits]: for all $x \in \mathbf{L}_\lambda$, there is in $\mathbf{L}_\lambda$ a surjection $\xi \to x$, where $\xi <\lambda$.

Note: Devlin shows in Ch. B.5 of the Handbook of Mathematical Logic: For every limit $\alpha$, there is a $\mathbf{\Sigma}_1(\mathbf{L}_\alpha)$ surjection $\alpha \to \mathbf{L}_\alpha$

I am trying to avoid using this. Arguments using admissible sets are most definitely ok.

A reference to a published proof would be excellent!

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    $\begingroup$ There is no reason this should be simple and avoid fine structure. $\endgroup$
    – Monroe Eskew
    Commented May 25, 2019 at 9:25
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    $\begingroup$ Other than showing that $L_\mu$ is an elementary submodel of some appropriate $L_{\kappa^+}$, which I don't even think is necessarily true, how do you propose to prove this statement? It's a strong statement. $\endgroup$
    – Asaf Karagila ♦
    Commented May 25, 2019 at 10:04

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Here is a possible route for a proof, but I'm not 100% sure if the idea holds up. I'm posting it here as CW so others can make adjustments if necessary.

Step 1: $\mu$ is an admissible ordinal.

This follows because if there was a $\Sigma_1(L_\mu)$ definition of a map from some $\gamma<\mu$ onto $L_\mu$ (or just onto $\mu$), then $L_\lambda$ wouldn't think that $\mu$ is a cardinal, let alone a successor cardinal.

Corollary 1: $L_\mu\models\sf KP$.

Step 2: $L_\lambda\models L_\mu\models\sf Replacement$.

Suppose that $\varphi$ defines a function $f\in L_\lambda$ from an element of $L_\mu$ to $L_\mu$, then since $L_\lambda$ satisfies that all the elements of $L_\mu$ have cardinality less than $\mu$, we get that in $L_\lambda$ this function must be bounded in $L_\mu$. In particular, we get full Replacement as far as $L_\lambda$ is concerned.

Corollary 2: $L_\mu\models\sf Replacement$.

And that's it. That's $\sf ZF^-$.

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    $\begingroup$ Alternatively to the first step, you can pretty easily prove Extensionality, Infinity, and Union from the definition of $L_\mu$ for a limit ordinal $\mu>\omega$. $\endgroup$
    – Asaf Karagila ♦
    Commented May 25, 2019 at 10:25
  • $\begingroup$ Thanks Asaf for your sussgestions. I have the following remarks: 1. Your Step 1 uses a result that's fairly involved, and for all I know uses Devlin's result which I was trying to avoid. $\endgroup$
    – Ramez L. Sami
    Commented May 26, 2019 at 1:07
  • $\begingroup$ You can use the suggestion in my comment, and just prove the rest of the axioms by hand. That's not very hard, actually. $\endgroup$
    – Asaf Karagila ♦
    Commented May 26, 2019 at 1:18
  • $\begingroup$ Thanks Asaf for your sussgestions. Sorry for my late reaction (life is harder, outside of mathematics...) I have the following remarks: 1. Your Step 1 uses a result that's fairly involved, and for all I know uses Devlin's result which I was trying to avoid. Do you know a direct proof of the result you are quoting? A reference? 2. Your Step 2 assumes that $\mu$ is regular in $\mathbf{L}_\lambda$. True, but requires an argument: essentially showing that if $\mu$ is the successor of $\kappa$ in $\mathbf{L}_\lambda$, then $\kappa$ is the largest cardinal in $\mathbf{L}_\mu$. $\endgroup$
    – Ramez L. Sami
    Commented May 26, 2019 at 1:23
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    $\begingroup$ Well. You said using admissible sets is okay. I'm not sure where's the problem. I'm just using the facts of $L$... $\endgroup$
    – Asaf Karagila ♦
    Commented May 26, 2019 at 8:01

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