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Nov
9
revised Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
argument below the theorem is not a proof of the theorem, just an example of some subtleties of the theorem, and the example (I hope) partially answers the question.
Nov
9
answered Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
Nov
8
comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
extension. These facts are all true at least in the case when a small forcing is used. There are large forcings which will destroy certain large cardinals, large forcings which will create so-called generic large cardinals, and, if I remember correctly, forcings that will resurrect large cardinals which were previously killed. There are many articles exhibiting this type of phenomenon, though I don't have any specific references at the moment.
Nov
8
comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
@JosephVanName. Your question is vague (to me at least) but now I think I get what you are after. The point of my first comment was that the consistency of an I0 gives an I1 with your embedding in a generic extension, similarly I1 suffices for an I3 with that property. These embedding already exist in the ground model and are preserved to the generic extension. This is one direction of the Levy-Solovay phenomenon. The other direction is that large cardinals are not created in the generic extension. So your desired $\lambda$, if it is not already I3, say, then it will not become I3 in the
Nov
8
comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
Maybe I've misunderstood your question. Are you asking for an inaccessible above an I3 which is itself not I3, that you can collapse and find a non-trivial embedding in the generic extension? Or do you want an inaccessible (or perhaps something stronger), not necessarily above an I3 that you can singularize through forcing and introduce an embedding in the generic extension? Or maybe something else altogether?
Nov
8
comment Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?
@JosephVanName. Don't I1 cardinals in V already imply the existence of many I3 cardinals below? If we use a small partial order, these I3 cardinals are preserved to the generic extension. More specifically, if $\lambda$ is I1, then there is a $\bar{\lambda}$ below it and a $j$ a witnessing that it is I3. If $G\subset P$ is generic and $|P|<crit(j)$, then $j$ extends in the generic extension to a non-trivial elementary embedding $k$ from $V_{\bar{\lambda}}^{V[G]}$ to itself. A similar argument gives the same for I1, assuming I0.
Oct
24
comment Can one take roots of rank-into-rank embeddings infinitely many times?
@Jospeh Van Name I have nothing to add, but I suspect Prof. Scott Cramer at Rutgers could share some insight link. His papers on Inverse Limits and the various reflections possible using them seems very relevant. ~~~~
Oct
24
awarded  Informed
May
17
comment Where does this strengthening of I1 stand?
@Joseph: Can I ask what the motivation is for this axiom? I think your subsets A cannot contain any member of the critical sequence, or maybe just \kappa, the critical point itself. Maybe there are other subsets that can't be preserved either?
Jan
2
comment A question on rank-to-rank embeddings
I'm a little confused by the remark on Skolem functions. Do you have an example of a total Skolem function $f$ whose image $j(f)$ is partial? Is your worry that such functions won't be total in general, or do you have a specific set A and/or a formula $\varphi$ in mind?
Jul
2
awarded  Curious
Jun
11
comment Are normal ultrafilters generated by conditional closure systems?
@JDH: If I'm thinking of the correct "Solovay's lemma" that was mentioned, I think there is a nice proof at Andres Caicedo's blog.
Jun
10
accepted Stationary sets in HOD
Jun
10
comment Stationary sets in HOD
Right. That was sloppy of me. I'll need to think about your great answer some more. In the meantime, I guess I'm trying to articulate a question about the definability (not necessarily using only ordinal parameters) of clubs and stationary sets.
Jun
10
comment Stationary sets in HOD
@JDH. Thank you for your answer. One interesting fact that I glean from it is that stationary sets are definable, whereas there are club sets that are not not (at least from ordinal parameters). Is this correct? Since you mention the forcing which kills a stationary/co-stationary set, is this the only way to conclude that there are club sets which are not definable, or is there an easier argument to see this?
Jun
10
asked Stationary sets in HOD
May
20
awarded  Fanatic
Feb
25
accepted Elementary Embeddings and Relative Constructibility
Feb
16
comment Antichains and the Knaster Property
@Paul McKenney: Thanks for your comment. I posed the question because I had a deep misunderstanding of the concept and you alerted me to this fact. Thanks!
Feb
16
revised Antichains and the Knaster Property
The background of the question was false and made for a confusing question.