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If $\kappa$ is almost huge (another large cardinal property), then for each family of size $\kappa$ of graphs of size $<\kappa$, one of the graphs embeds into another one from the family. See this p …

First of all, most people would be rather unhappy with a chain like yours of countable models of ZFC. You can do mathematics in these countable models, but it is still mind bending. An inaccessible c …

For $\omega_1$ the argument is as follows: Consider the filter generated by all closed and unbounded subsets of $\omega_1$. Under AD, this filter is an ultrafilter (this is due to Solovay). The filt …

Just one clarification to Guillaume's answer (his answer has been edited by now): Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The rela …

I just this minute read in a paper by Sharon and Viale that Schimmerling has shown the failure of $\square(\aleph_n)$ for two consecutive $n$ implies projective determinacy and hence the consistency o …

Every inaccessible cardinal is a fixed point of the operation $P$ that assigns to every set $X$ of ordinals the set $P(X)=\{2^{|\alpha|}:\alpha\in X\}\cup\bigcup X$. On the other hand, every (nonempty …

There are two ingredients in your question: First, statements invariant under forcing and second, large cardinals. The existence of certain large cardinals is a strengthening of ZFC. Somewhat surpri …

Matousek showed that for every $K\gneq 1$ every infinite metric space $X$ has an infinite subspace that either embeds into the real line by a $K$-bi-Lipschitz function or in which the distances of any …

First of all, why is it so unreasonable to think that ZFC itself is contradictory? Because we have a good intuition about sets and we have a lot of experience with ZFC. The same basically applies to …

As was pointed out in some answers to this question, since the large cardinal axiom are linearly ordered by consistency strength, there is a natural direction in which we can strengthen set theory. Si …