First Post

Well... This is my first post on this blog, and I have absolutely no idea how to start it.

Should I make it about myself? about my life? about my academic status? How I about I tell cool stories from my life, perhaps inebriated adventures? army experiences? Maybe I should write about mathematics. Perhaps some nice proof or some nice theorem?

Well, it seems that in a most self-referential way I already began my first post and I am writing about it in itself. Now that we have done that, let me do all of the above.

My name is Asaf Karagila. I live in Israel, and have lived here my whole life. I just finished my M.Sc. in Ben-Gurion University under the supervision of Uri Abraham. My thesis was titled Vector Spaces and Antichains of Cardinals in Models of Set Theory. It extends a classical construction by Läuchli of a vector space which has no basis. I also presented a paper by Feldman and Orhon in which they generalize Hartogs theorem, I slightly improve upon their original proof and I extend their results.

Now, and for the next couple of years, I am a Ph.D. student in the Hebrew University in Jerusalem, supervised by Menachem Magidor. There is no concrete topic right now, and I going to have a busy year full of studying. It's a good thing, I hope.

All those things will find their way in details to this site (as well as my thesis itself once I am finished with minor corrections and improvements).

To finish, here is a bit of mathematics.

-------\(\newcommand{\Seq}{\operatorname{Seq}}\)

Definition. We say that \(A\) is Dedekind-finite if \(B\subseteq A\) and \(|A|=|B|\) then \(A=B\).

Proposition.\(A\) is Dedekind-finite if and only if \(\aleph_0\nleq|A|\) Proof. Exercise.

Assuming the axiom of choice the above proposition shows that a set is Dedekind-finite if and only if it is finite (i.e. has an injective function into a finite ordinal). However without the axiom of choice it is consistent to have infinite Dedekind-finite sets.

For a set \(A\), let \(\Seq(A)=\{f\colon n\to A\mid f\text{ injective}\}\).

Lemma. Let \(A\) be a Dedekind-finite set, then \(\Seq(A)\) is also Dedekind-finite. Proof. If \(\Seq(A)\) was not Dedekind-finite it would have a countably infinite subset \(B\). Each sequence in \(B\) defines a finite subset of \(A\), their union is a union of enumerated sets and therefore can be well-ordered. We assumed that \(B\) is infinite and therefore the union must be infinite, since a finite set can only have finitely many different enumerations.

Therefore \(\bigcup_{f\in B}\rng(f)\) is a countably infinite subset of \(A\), in contradiction to the assumption that \(A\) is Dedekind-finite. \(\square\)

Theorem. (Tarski) Suppose that there exists an infinite Dedekind-finite set, then there exists a set \(\cal A\) of Dedekind-finite sets such that \((\cal A,\subseteq)\) is order isomorphic to \((\mathbb R,\lt)\).

Proof. Let \(A\) be an infinite Dedekind-finite set, and let \(S=\Seq(A)\). Let \(\{A_r\mid r\in\mathbb R\}\subseteq\power(\omega)\) be a chain of order-type \((\mathbb R,\lt)\) (e.g., fix an enumeration of the rationals, and use it to define \(A_r\) as the indices of the rationals in the Dedekind cut of \(r\)). Let \(S_r=\{f\in S\mid |f|\in A_r\}\), and \(\mathcal A=\{S_r\mid r\in\mathbb R\}\). If \(r\lt t\) then \(A_r\subsetneq A_t\) and therefore \(S_r\subsetneq S_t\), and since \(S\) is Dedekind-finite we have that \(S_r\) is Dedekind-finite as well. Furthermore, by Dedekind-finiteness if \(r\neq t\) we have that \(|S_r|\neq|S_t|\). \(\square\)

Note that the axiom of choice was not used anywhere in the above proofs. The lemma used the fact that we took a countable union over enumerated sets; and the theorem used the fact that we can find a chain of order-type \((\mathbb R,\lt)\) in \(\power(\omega)\) which is again a choice-free proof.

Corollaries. Suppose that there exists an infinite Dedekind-finite set, then

1. there are at least \(2^{\aleph_0}\) Dedekind-finite sets of different cardinalities;
2. there is an infinite descending sequence of cardinals.

Ah, how strange can be the universe when the axiom of choice is absent!