A (fairly) interesting equivalent to the negation of the continuum hypothesis

I recently saw an interesting equivalent statement to the negation of the continuum hypothesis. The statement is essentially Freiling’s axiom of symmetry, presented in a very slightly different manner. It’s interesting in that it seems more geometric and less abstract than the continuum hypothesis itself. In this post, I briefly rehash the definition of the continuum hypothesis and prove the claimed equivalence.

The continuum hypothesis (CH) is the claim that 2^{\aleph_0} = \aleph_1, or equivalently that there are no cardinalities intermediate between the cardinality of the naturals and the cardinality of the reals. It’s well known that this is independent of ZFC, and this is often taken to mean that its truthhood is “unknowable”, or that we are free to assume CH or its negation as we please. In a sense this is true. However, it is worth stressing that there is nothing inherently special about the ZFC axioms. We have good reasons for believing that they are consistent and they are powerful enough to formalise most of mathematics in, but just as choice and replacement were added for pratical purposes it is reasonable to suppose that we might produce other useful axioms that we wish to add that appear “plausible”. The study of large cardinal axioms is a well established branch of set theory that considers various strong axioms that could be added to ZFC (“strong” in the sense of consistency strength; all large cardinal axioms prove Con(ZFC)) that are motivated by various heuristics or desired consequences. However, none of the current large cardinal axioms settle the continuum hypothesis.

We now introduce the equivalent statement to not CH. The statement seems to me to be very intuitive, but I’m not sure if this is really just because my intuition of how infinite sets behave is a bit suspect (I am also somewhat drawn to the V=L axiom, and clearly I’m not allowed both!).

Denote the unit interval [0,1] by I. Given a set S \subset I \times I and x \in I, define S_x to be the set of all (y,x) \in S, i.e. S_x is the “row” of S at height x. We say that S is thin if for every x \in I, S_x is countable. Given S \subset I \times I, define the reflection of S, denoted \bar{S}, to be the set of all (x,y) such that (y,x) \in S (i.e. the reflection of S along the line y=x). Intuitively, it seems that taking the union of any thin set S with its reflection should never give us all of I \times I; thin sets “don’t fill up very much of I \times I“. In fact, it’s easy to show that this claim is equivalent to the continuum hypothesis being false.

The negation of the continuum hypothesis is equivalent to there being no thin set S with S \cup \bar{S} = I \times I.

Suppose that CH is true and let \pi: I \rightarrow \omega_1 be a bijection. Define a set S as follows: for each x \in I, define S_x to be the set of all (y,x) such that \pi(y) \leq \pi(x). Each S_x is clearly countable (by definition of \omega_1!) and S \cup \bar{S} = I \times I.

Now suppose that CH is not true and let S be any thin subset of I \times I. Let Y \subset I be a set of cardinality \aleph_1. Let X = \{x | (x,y) \in S \wedge y \in Y\}. |X \cup Y| = \aleph_1 so there exists x \in I such that x \notin X \cup Y. As S_x is countable there exists some y \in Y \setminus S_x. So (x,y) \notin S by our choice of x and (y,x) \notin S by our choice of y.

This entry was posted in Uncategorized. Bookmark the permalink.

One Response to A (fairly) interesting equivalent to the negation of the continuum hypothesis

  1. John Baez says:

    There’s some discussion of your post and also your use of the word “truthhood” on Google Plus, initiated by Richard Elwes.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s