The following is stated on Wikipedia for Mahlo cardinals. Unfortunately, it's not sourced. Where can I find details? I wasn't able to google any articles dealing with Mahlo cardinals in $L$.

Since $On⊂L⊆V$, properties of ordinals that depend on the absence of a function or other structure (i.e. $\Pi_1^{ZF}$ formulas) are preserved when going down from $V$ to $L$. Hence initial ordinals of cardinals remain initial in L. Regular ordinals remain regular in $L$. Weak limit cardinals become strong limit cardinals in $L$ because the generalized continuum hypothesis holds in $L$. Weakly inaccessible cardinals become strongly inaccessible. Weakly Mahlo cardinals become strongly Mahlo. And more generally, any large cardinal property weaker than 0# (see the list of large cardinal properties) will be retained in $L$.

  • $\begingroup$ Did you check Jech or Kanamori? Probably Drake has a proof of that too. $\endgroup$ – Asaf Karagila May 19 '16 at 13:19
  • $\begingroup$ I've checked Jech and Kanamori, but I was looking for Mahlo in the register, it may be proven via Weakly Compact cardinals and then just by the way mentioning, that a weakly compact cardinal is also Mahlo. I would prefer a direct proof for Mahlo cardinals, do you think it's done that way somewhere? $\endgroup$ – mikulas May 19 '16 at 20:18

Let $\kappa$ be a Mahlo cardinal in $V$, i.e. let $\kappa$ be inaccessible such that $S:= \{ \alpha \in \kappa \mid \alpha \text{ is regular} \}$ is stationary in $\kappa$.

First, note that $L \models \kappa \text{ is inaccessible}$. Indeed, if $L \models \kappa \text{ is not a cardinal}$, then there is some $\mu < \kappa$ and some $f \in L$ such that $L \models f \colon \mu \to \kappa \text{ is surjective}$. However, this is a $\Sigma_{0}$ property and hence, in $V$, $f \colon \mu \to \kappa$ is surjective. Contradiction. Since $\kappa > \omega$ (as an ordinal), this also yields that $L \models \kappa \text{ is uncountable}$. Repeating this argument with cofinal $f \colon \mu \to \kappa$ yields that $L \models \kappa \text{ is regular}$.

Since $L \models \operatorname{GCH}$, it now suffices to prove that $L \models \kappa \text{ is a limit cardinal}$. This is trivial, because for any ordinal $\gamma < \kappa$, we have that $(\gamma^{+})^V < \kappa$ and since cardinals in $V$ are cardinals in $L$, this proves $$L \models \forall \gamma < \kappa \exists \gamma < \mu < \kappa \colon \mu \text{ is a cardinal}.$$

Now let $T := \{ \alpha \in \kappa \mid L \models \alpha \text{ is regular} \}$. By the argument given above, any $\alpha$ that is regular in $V$ is regular in $L$ and hence $S \subseteq T$. Suppose that $L \models \kappa \text{ is not Mahlo}$. Then there is some $C \subseteq \kappa$ such that $L \models C \text{ is club in } \kappa \text{ and } C \cap T = \emptyset$. Being club is a $\Sigma_0$, property and hence $C \subseteq \kappa$ is club in $\kappa$. Since $S \subseteq T$, we have that $C \cap S = \emptyset$ and hence $S$ is not stationary in $V$. This is a contradiction and we therefore must have that $L \models T \text{ is stationary}$.

Thus, $\kappa$ remains Mahlo in $L$.


Let $C(k)$ be the set of club subsets of $k.$ Let $R(k)=\{l\in k: l=cf(l)\}.$ Observe that for any set $S\subset On,$ if $S\in L$ then

(1) $\forall a\in On \;[\;\{b\cap S :b\in a\} \in L\;], \; \text {and}$

(2) $\forall a\in On\;[\;a=\cup (a\cap S)\iff L\Vdash (a= \cup (a\cap S)\;].$

(3) Also observe that $\forall a\in On\;[a=|a|\implies L\Vdash a=|a|.)]$

Let $C(k)$ be the set of club subsets of $k.$ From (1) and (2) we have $ C(k)\supset (C(k))^L.$ Let $R(k)=\{l<k: l=|l|\}.$ From (3), we have $R(k)\subset (R(k))^L.$

So for any $c\in (C(k))^L$ we have $c\in C(k), $ so $\emptyset\ne c\cap R(k)\subset c\cap (R(k))^L.$

Remark. If $k=|k|>\omega$ and $R(k)$ is stationary in $k,$ then $k$ must be weakly inaccessible. Obviously $k$ can't be a successor cardinal. If $k$ is singular then

(i): If $k>cf(k)=\omega $ let $f(n):\omega \to k$ be a co-final strictly increasing map with $f(0)=\omega.$ Let $g(n)=f(n)+1$ for $n\in \omega.$ Then $\{g(n):n\in \omega\}$ is club in $k$ and contains no cardinals.

(ii): If $k>cf (k)=l>\omega,$ let $S\subset k$ with $|S|=l$ and $\cup S=k.$ Let $C=\{a\in k: l<a=\cup (a\cap S)\}.$ It is easy to show that $C$ is club in $k.$ For $a\in C$ we have $cf (a)\leq |a\cap S|\leq l<a.$ So $C$ has no regular members.


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