Doesn't the unprovability of the continuum hypothesis prove the continuum hypothesis?

Question

The Continuum Hypothesis says that there is no set with cardinality between that of the reals and the natural numbers. Apparently, the Continuum Hypothesis can't be proved or disproved using the standard axioms of set theory.

In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals. It was proven that the CH can't be disproven. Equivalently, it was proven that one cannot construct a counterexample for the CH. Doesn't this prove it?

Of course, the issue is that it was also proven that it can't be proved. I don't know the details of this unprovability proof, but how can it avoid a contradiction? I understand the idea of something being independent of the axioms, I just don't see how if there is provably no counterexample the hypothesis isn't immediately true, since it basically just says a counterexample doesn't exist.

I'm sure I'm making some horrible logical error here, but I'm not sure what it is.

So my question is this: what is the flaw in my argument? Is it a logical error, or a gross misunderstanding of the unprovability proof in question? Or something else entirely?

Answer 1

I think the basic problem is in your statement that "In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals." Actually, to disprove CH by this strategy, one would have to produce a counterexample and prove that it actually has cardinality between those of $\mathbb N$ and $\mathbb R$.

So, from the fact that CH can't be disproved in ZFC, you can't infer that there is no counterexample but only that no set can be proved in ZFC to be a counterexample.

Answer 2

How would you prove that not all groups are abelian? Namely, how would you prove that the axioms of groups do not imply $\forall x\forall y(xy=yx)$?

Well. You'd find a model for the theory, namely a group, which is not abelian. At the same time, you can also show this is not disprovable by finding an abelian group.

So, does that mean that all groups are abelian, because the axiom $\forall x\forall y(xy=yx)$ is independent of the axioms of a group? No. It does not.

The independence of the continuum hypothesis from ZFC goes along the same lines. We can show there are models of ZFC where CH holds, and others where it fails. This is regardless to the techniques we use, which are clever and useful for so much more in set theory.

Answer 3

The situation is really rather similar to that of non-Euclidean geometry.

From the first four Euclidean postulates, you can neither prove nor disprove the fifth (parallel) postulate. This has been proven only two millennia after Euclid wrote his Elements. If you assume the fifth postulate is true, you get Euclidean geometry; if you assume it is false, you get elliptic or hyperbolic geometry. The common ground to both is absolute geometry.

Similarly, from ZFC, you can neither prove nor disprove the continuum hypothesis. If you assume it is true, you get one kind of set theory; if you assume it is false, you get another. The common ground to both is ZFC.

Answer 4

Here's an example axiomatic system:

1. There exist exactly three objects $A, B, C$.
2. Each of these objects is either a banana, a strawberry or an orange.
3. There exists at least one strawberry.

Let's name the system $X$.

Vincent's Continuum Hypothesis (VCH): Every object is either a banana or a strawberry (i.e., there are no oranges).

Now, to disprove this in $X$, you would have to show that one of $A, B, C$ is an orange ("construct a counterexample"). But this does not follow from $X$, because the following model is consistent with $X$: A and B are bananas, C is a strawberry.

On the other hand, VCH does not follow from $X$ either, because the following model is consistent with $X$: A is a banana, B is a strawberry, C is an orange.

As you can see, there is no contradiction, because you have to take into account different models of the axiomatic system.

Answer 5

CH: "There is no set with cardinality between that of the reals and the natural numbers."

disprove CH by: demonstrating such a set

You can't demonstrate such a set. Therefore, you can't disprove it.

To prove it, you have to show that ALL sets have cardinality:

1. less than or equal to natural numbers -or-
2. greater than or equal to real numbers

You can't prove it either.

Therefore, you can neither prove nor disprove CH.

The basic problem is the unfounded assumption that you can prove something by simply showing that you can't disprove it. You'll have to prove that one first.

Actually, the continuum hypothesis itself demonstrates that you can't, by virtue of "you can't prove it" above.