# 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 a Dedekind-infinite set, then

1. there are at least $2^{\aleph_0}$ Dedekind-finite 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!

### 16 Responses to First Post

1. Welcome, Asaf! It’s great to have you!

• Asaf Karagila

Thank you Peter, it is great to be here!

2. Welcome to Boolesrings, Asaf! And what a very nice bit a mathematics for your first post…

• Asaf Karagila

Thank you Joel!

3. Welcome Asaf! I look forward to reading more of your posts!

4. Victoria Gitman

Welcome Asaf! Thanks for the great first post! I always wanted to learn about Dedekind-finite sets.

• I guess you’re in luck. I foresee several posts deal with them. :-)

5. Micheal Pawliuk

Neat. Good to have you on Boolesrings!

It might be worth pointing out what this has to do with AC, and that none of your proofs use AC.

6. Are special requests accepted? I would love to hear more about your extension of Läuchli’s construction. Moreover, since Läuchli’s construction is a periodic topic on MathOverflow, it would be nice to point people to a nice presentation in addition to the original paper in German.

• Francois, of course I will make such post. I also intend to write the details into a paper, and after two minor improvements on that very chapter my thesis will also find its way onto this site. All in due time. To be perfectly honest, my work began as an answer on MSE and an answer on MO. Both, however, used atoms, whereas my thesis used a direct forcing construction.

7. Hi Asaf, nice to see you in the sphere of mathematicians with an internet presence!

• Thanks David! I think that I have an internet presence ever since I started being [really] active on MSE, but now I have a small place where I can complain about all that choice the modern world is giving us and how to destroy this choice with force[ing]!

• Ah, that is true. But having a blog is a bit different. It’s your own turf, rather than being known to hang out in a certain playground (MO is also a playground, by this analogy!)