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This is a follow up to my previous question on Silver indiscernibles.

Background: Suppose that $0^\#$ exists, $\alpha<\lambda$ are limit ordinals, $i_\alpha$ is the $\alpha$th Silver indiscernible, and $\lambda$ is contained in $L_{i_\alpha}$ (which is just the model generated by the first $\alpha$ indiscernibles). In $V$, we can define an injection from $\lambda$ to $\omega\times\alpha^{< \omega}$ by simply specifying which term and indiscernibles generate each member of $\lambda$.

Question: Suppose that $\lambda$ is not a Silver indiscernible, is there also a constructible injection from $\lambda$ to $\omega\times\alpha^{< \omega}$?

Edit: By Francois' answer to my previous question, I can now see that the injection described in the background can't be in L (as the union of members in the range of the function will give an infinite constructible set of indiscernibles), so I just want to emphasize that any constructible injection to $\omega\times\alpha^{< \omega}$ (or just to $\alpha$) would be fine as an answer.

Edit #2: Due to the answer written by Apostolos, I've added the assumption that $\lambda$ is not a Silver indiscernible.

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Your LaTeX skills improve, but you have much to learn young grasshopper :-) – Asaf Karagila Jul 18 '11 at 22:58

In $L$ every Silver indiscernible is a cardinal (since every real uncountable cardinal is a Silver indiscernible). If $\alpha$ is a limit ordinal and $i_\alpha>\alpha$ then since the indiscernibles are unbounded in $L_{i_\alpha}$ we have that there is some $\beta<\alpha$ such that $i_\beta>\alpha$.This $i_\beta$ is a cardinal in $L$ so there can be no constructible injection from $i_\beta$ to $\alpha$. Because $L$ satisfies the Axiom of Choice, we have that $|\omega\times\alpha^{<\omega}|=|\alpha|$ in $L$ and thus the injection you are looking for doesn't exist.

EDIT: It is not clear what exactly you are looking for with your edit. Obviously, for any ordinal larger than the $i_\beta$ that I used above, (for example $i_\beta\cup\{i_\beta\}$ which as a successor ordinal is not a Silver indiscernible) no such bijection exists. Of course there are many ordinals to which a bijection from $\alpha$ exists (namely $(|\alpha|^+)^L$ many), but (the point of my post is that) in the case when the $\alpha$th Silver indiscernible is not a fixed point of the ordinals, there are ordinals for which no bijection between them and $\alpha$ exists.

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Perhaps I'm missing something, but in order to complete this argument, you need to show that $i_\beta$ < $\lambda$, and it's not clear that there is an indiscernible between $\alpha$ and $\lambda$. – ZeroDagger Jul 19 '11 at 10:53
@ZeroDagger: I took as $\lambda$ the indiscernible $i_\beta$. – Apostolos Jul 19 '11 at 11:31
Thanks. Unfortunately, it seems that I forgot to add the assumption that $\lambda$ is not a Silver indiscernible. I've now edited my question accordingly. – ZeroDagger Jul 19 '11 at 11:41
Apostolos has the right idea. Every Silver indiscernible is in fact inaccessible in $L$. There will be a constructible injection $\lambda \to \omega\times\alpha^{<\omega}$ if and only if $\lambda < \max((|\alpha|^+)^L,\omega_1^L)$. If $\alpha < i_\alpha$, then $i_\alpha$ is necessarily much larger than $\max((|\alpha|^+)^L,\omega_1^L)$. – François G. Dorais Jul 19 '11 at 12:50

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