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Stefan Mesken's user avatar
Stefan Mesken's user avatar
Stefan Mesken
  • Member for 10 years, 4 months
  • Last seen more than 4 years ago
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awarded
 Nice Answer
comment
Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?
The link in my previous comment no longer works. Here is an updated link to Gabe's talk.
awarded
 Yearling
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Explicit, small resolving sets for Hamming graphs
@kodlu $\infty$ is included since the graph may not be connected. And one relevant paper is chapter 6 of ON THE METRIC DIMENSION OFCARTESIAN PRODUCTS OF GRAPHS.
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Explicit, small resolving sets for Hamming graphs
@Bullet51 Any resolving set of small size that you can make available to me (so that I can actually use it in an algorithm) counts as 'explicit'.
revised
Explicit, small resolving sets for Hamming graphs
added 153 characters in body
revised
Explicit, small resolving sets for Hamming graphs
edited body
asked
Explicit, small resolving sets for Hamming graphs
awarded
 Critic
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Amalgamation via elementary embeddings
@MonroeEskew My previous comment meant to say that Andreas' answer does indeed give a positive answer to your question, at least if we also assume that $V$ satisfies the ultrapower axiom.
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Amalgamation via elementary embeddings
If $V$ is an extender model (it is enough to assume that $V$ satisfies the ultrapower axiom), then this idea of amalgamating $M_0, M_1$ works and the diagram commutes.
awarded
 Excavator
revised
V=L and a Well-Ordering of the Reals
improved formatting and fixed two typos
suggested
Approve
V=L and a Well-Ordering of the Reals
awarded
 Yearling
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The complexity of sorting a list having one free cell
To clarify: A move consists of switching an occupied row with the empty row?
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Function $f:\kappa\to\alpha$ with small fibers where $\alpha\in\kappa$
@Qfwfq You're welcome.
answered
Function $f:\kappa\to\alpha$ with small fibers where $\alpha\in\kappa$
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Function $f:\kappa\to\alpha$ with small fibers where $\alpha\in\kappa$
@Qfwfq How do you prove $|\coprod_{x\in \alpha}f^{-1}(\{x\})|\leq\alpha\beta$ in $\mathrm{ZF}$ if you don't have a choice function $x \mapsto i_x \colon f^{-1}(\{x\}) \to \beta$?
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Does $0^{\#}$ imply that $L$'s set generic multiverse has irreconcilable theories?
@Joel I've thought about that as well. Unfortunately $V$ and $L$ massively disagree about what $\mathrm{Coll}(\kappa, \kappa^{+})$ is... I've also thought about reflecting a possible failure via (very) large cardinals and then use something along the lines of proposition 3. But thus far I don't see how to do it.
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