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According to Wikipedia, if $0^{\sharp}$ exists, then every uncountable cardinal in $V$ satisfies every large cardinal property in $L$ that can be realized in $L$, e.g. weak compactness, total ineffability, etc. It's easy enough to see why every uncountable in $V$ will be inaccessible, or even Mahlo, in $L$.

How can one see that some of the slightly larger large cardinal properties (e.g. weak compactness, total ineffability, etc.) are satisfied in $L$ by the uncountable cardinals in $V$? Is there a good reference for some of these results?

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There is a very nice section about $O#$ in Kanamori's book "The Higher Infinite". And I think the questions you ask are answered in that section or somewhere within that text. – Michael Blackmon Apr 15 2011 at 2:42

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If $0^\sharp$ exists, then every uncountable cardinal $\kappa$ of $V$ is one of the Silver indiscernibles in $L$, and this implies that $L_\kappa$ is an elementary substructure of $L$. This implies that $\kappa$ is a limit cardinal in $L$ and therefore, since some of the indiscernibles are regular, that $\kappa$ is inaccessible in $L$. Every order-preserving map on the indiscernibles induces an elementary embedding $j:L\to L$, and thus every indiscernible is the critical point of such a $j$. From this, it follows that every such $\kappa$ has the tree property in $L$, because if $T$ is any $\kappa$-tree in $L$, then $j(T)$ has nodes on the $\kappa$-th level, which gives you a $\kappa$-branch in $T$. But being inaccessible and having the tree property is equivalent to being weakly compact, so every such $\kappa$ is weakly compact in $L$. You can get other properties by arguing with the embedding like this.

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Thanks Joel! I look forward to meeting you in a week or two at Fields. – Amit Kumar Gupta Apr 15 2011 at 14:23
Oh, yes, I'll be coming up to Toronto soon. – Joel David Hamkins Apr 15 2011 at 23:25

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