# I do not understand how the continuum hypothesis can be undecidable

I have been thinking about what it means for the continuum hypothesis to be undecidable.

In short, here is my understanding

$$\aleph_{0}$$ is the cardinality of the set of integers.

$$\aleph_{1}$$ is the cardinality of the smallest set bigger than $$\aleph_{0}$$

According to the continuum hypothesis, $$2^{\aleph_{0}}=\aleph_{1}$$

Now, here is why I do not understand:

Let us suppose there exists some set S which is greater than $$\aleph_{0}$$ but smaller than $$2^{\aleph_{0}}$$

Then, it will be easy to prove by Cantor's diagonal argument that S violates the hypotheses. So, if the continuum hypothesis is false, it must be provably false.

This is imposible if we assume that the continuum hypothesis is undecidable, as it means that the truth of the continuum hypothesis is consistent with ZFC, and there by it cannot be provably false.

Therefore, since if it was false it would be PROVABLY false, which is impossible, the continuum hypothesis is true.

Is there a problem with this reasoning? Have I tripped over my own argument, or is there no reason to assume it is provably false? Please explain.

• Why is it provably false if it is false in some model of ZFC? – Asaf Karagila Apr 2 at 7:12
• Godel proved, by using his "constructible class" $L$ that if the axiom system ZF is consistent then so is ZF+AC+GCH. (AC is the Axiom of Choice and GCH is the Generalized Continuum Hypothsis.) Paul Cohen invented Forcing and used it to prove that if ZFC (i.e.ZF+AC) is consistent then so are ZF+($\neg$ AC) and $ZFC+(\neg CH)$.... So if ZFC is consistent then neither CH nor $\neg$CH is a theorem of ZFC. – DanielWainfleet Apr 3 at 6:13

Yes, if CH is false, there is some $$S\subseteq\Bbb R$$ such that $$\aleph_0<|S|=\aleph_1<2^{\aleph_0}$$. But in different models of ZFC this $$S$$ might be very different. It is not necessarily that the same definition always works as a counterexample to CH.