## A Practical Guide to Using Countable Elementary Submodels

So, you may have heard about these things called countable elementary submodels. You may have heard that they work like magic and do all sorts of amazing things. “Mathematical voodoo” some might say. “Witchcraft!” others declare. Hearing this you become intrigued and set out to harness this black power. You quickly realize that there are very few places to learn this dark art; the protectors of this knowledge don’t want it leaking out.

Here I hope to lay out the essential things you need to know (and omit the things you don’t need to know) so that you can start using countable elementary submodels. I am going to lay out as little of the machinery as possible and display only the relevant applicable facts you will need for most proofs involving elementary submodels.

1. A Countable Submodel of What?

The universe of all sets $V$ is a ‘model’ for set theory, but it is too big. If we did have a model for set theory we would know that there is a countable submodel of it, by Lowenheim-Skolem. Of course we can’t assert that set theory has a model as this would be equivalent to asserting the consistency of set theory. The clever way around this is to realize that any proof in mathematics only ever uses finitely many axioms of set theory and references only finitely many specific sets. It is always possible to find a model $H$ of those finitely many axioms and special sets. (Aside, for those of you who have seen this before, why doesn’t this violate the compactness theorem? It’s tricky.) Here $H$ will be our copy of the universe, just for a given proof, and we will take a countable submodel of $H$, not $V$. This is where the language “Take a large enough fragment of ZFC” comes from.

As it turns out there is a class of sets that we usually draw $H$ from. We usually take $H$ to be a set $H(\alpha)$, where $\alpha$ is a cardinal and $H(\alpha)$ is the set of all sets hereditarily of cardinality less than $\alpha$. This doesn’t really matter at all. So don’t fret about this. Continue reading A Practical Guide to Using Countable Elementary Submodels

## Secret Santa 3: The Paradox.

Last time I discussed the solution to Sam’s problem:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

I established, by induction, that it is impossible to do this. Great. We write “QED” and move on. There is a very convincing counter-argument that was brought to my attention by Jacob Tsimerman and a student at the Winter Canadian IMO camp. They proposed a method that seems like it should solve Sam’s problem in the positive. What exactly is going on in their method? Where is the mistake?

Jacob’s method. Player 1 chooses a large enough number N (say greater than 100); this is now their number. Player 1 writes down the numbers N+1 and N-1 on different pieces of paper and presents them face-down to Player 2. Player 2 chooses one of them and burns the other one without looking at it. The number Player 2 sees is their number.

## The Secret Santa Problem (Part 2)

Last time, just in time for Christmas, we looked at the Secret Santa Problem. Basically the problem is to set up a secret santa type gift exchange without using any external aids like random number generators. A similar problem given to me by Sam Coskey is the following:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

When Sam gave the problem to me he intended that each player would choose a natural number and then they would sequentially ask questions to each other, possibly refining their original numbers. In this sense it is more like a game of Guess Who than secret santa.

After much thinking, it turns out that there is a fairly easy solution to this problem.

Part 1. Can either player choose the number 0?

Well no, because that person would know that they have a smaller number than the other player.

Part 2. If both players know that $1, 2, \dots, k$ cannot be chosen then $k+1$ cannot be chosen (as $k+1$ would have to be the smaller of the two numbers). So by induction, no number can be chosen by either player.

The lesson here is that induction is a very useful technique! This sounds naive but, when problem solving for contests, induction is often overlooked. Here is another related problem:

Father/Son problem. Is it possible for two players to each choose a human being so that the two humans are father and son, but neither player knows who has the father and who has the son?

The solution is again fairly simple, and uses induction. This time we need a different type of induction. We observe that no player can pick the first ever human being (as they would have the father of the other player’s choice). Now if there is a set S of humans that cannot be chosen, then the sons of people in S cannot be chosen either.

There you go, induction wins again!