Aleph 0 as a large cardinal - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:46:30Z http://mathoverflow.net/feeds/question/45448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45448/aleph-0-as-a-large-cardinal Aleph 0 as a large cardinal arsmath 2010-11-09T15:44:45Z 2011-12-31T22:35:48Z <p>The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph_0$, and which are not?</p> <p>There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?)</p> <p>EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting.</p> http://mathoverflow.net/questions/45448/aleph-0-as-a-large-cardinal/45532#45532 Answer by Amit Kumar Gupta for Aleph 0 as a large cardinal Amit Kumar Gupta 2010-11-10T07:37:10Z 2010-11-10T07:37:10Z <p>This probably isn't the kind of thing anyone just knows off hand, so anyone who's going to answer the question is just going to look at a list of large cardinal axioms and their definitions, and try to see which ones are satisfied by $\aleph _0$ and which aren't. You could've probably done this just as well as I could have, but I decided I'd do it just for the heck of it. </p> <p>First of all, this doesn't cover all large cardinal axioms. Second, many large cardinal axioms have different formulations which end up being equivalent for uncountable cardinals, or perhaps inaccessible cardinals, but may end up inequivalent in the context of $\aleph _0$. So even for the large cardinals that I'll look at, I might not look at all possible formulations. </p> <ol> <li><strong>weakly inaccessible</strong> - yes, obviously</li> <li><strong>inaccessible</strong> - yes, obviously</li> <li><em>Mahlo</em> - no, since the only finite "inaccessibles" are 0, 1, and 2 as noted by Michael Hardy in the comments, and the only stationary subsets of $\omega$ are the cofinite ones</li> <li>weakly compact: <ul> <li><strong>in the sense of the Weak Compactness Theorem</strong> - yes, by the Compactness Theorem</li> <li><strong>in the sense of being inaccessible and having the tree property</strong> - yes, by Konig's Lemma</li> <li><em>in the sense of <code>$\Pi ^1 _1$</code>-indescribability</em> - no, it's not even <code>$\Pi ^0 _2$</code>-indescribable as witnessed by the sentence $\forall x \exists y (x \in y)$</li> </ul></li> <li><em>indescribable</em> - no, since it's not even <code>$\Pi ^0 _2$</code>-indescribable</li> <li><em>Jonsson</em> - no, the algebra $(\omega, n \mapsto n \dot{-} 1)$ has no proper infinite subalgebra</li> <li><em>Ramsey</em> - no, the function $F : [\omega ]^{&lt; \omega} \to \omega$ defined by $F(x) = 1$ if $|x| \in x$ and $0$ otherwise has no infinite homogeneous set</li> <li>measurable: <ul> <li><strong>in the sense of ultrafilters</strong> - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition, i.e. closed under finite intersections</li> <li><em>in the sense of elementary embeddings</em> - no, obviously</li> </ul></li> <li><em>strong</em> - no, obviously (taking the elementary embedding definition)</li> <li><em>Woodin</em> - ditto</li> <li>strongly compact: <ul> <li><strong>in the sense of the Compactness Theorem</strong> - yes, by the Compactness Theorem</li> <li><strong>in the sense of complete ultrafilters</strong> - yes, as in the case of measurables</li> <li><strong>in the sense of fine measures</strong> - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition</li> </ul></li> <li>supercompact: <ul> <li><em>in the sense of normal measures</em> - no, if $x \subset \lambda$ is finite and $X$ is the collection of all finite subsets of $\lambda$ which contain $x$, then the function $f : X \to \lambda$ defined by $f(y) = \max (y)$ is regressive, but for any $Y \subset X$, if $f$ is constant on $Y$ with value $\alpha$, then Y avoids the collection of finite subsets of $\lambda$ which contain <code>$\{ \alpha + 1\}$</code> and hence Y cannot belong to any normal measure on $P_{\omega }(\lambda)$</li> <li><em>in the sense of elementary embeddings</em> - no, obviously</li> </ul></li> <li><em>Vopenka</em> - no, take models of the empty language of different (finite) sizes</li> <li><em>huge</em> - no, obviously (taking the elementary embedding definition)</li> </ol>