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How do we know an $ \aleph_1 $ exists at all?

Here, you can see that $\aleph_2≦2^{\aleph_0}$, which is a contradiction to CH in ZFC. So if 'true statements in ZF is true in ZFC' is true, CH must be false in ZFC, which is a contradiction.

So does there exist a statement that is true in ZF but false in ZFC? Or am i getting it wrong?

I think this is really important since if the suitable Axiom (weaker than ZFC, stronger than ZF) is made in future, this makes us possible to apply all theorems in ZF to that axiomatic system.

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Where can we see that? If ZF could prove something false in ZFC, ZFC would be inconsistent (which is not impossible, but unlikely to be so easy to see ;) ). – tomasz Jul 30 '12 at 12:02
@tomasz: Godel's theorem says that if ZF does not have contradictions then ZFC does not have contradictions. Namely, if we proved something is false in ZF, we cannot prove it is true in ZFC unless ZF itself already contains a contradiction. – Asaf Karagila Jul 30 '12 at 12:04
@AsafKaragila: Yes. That only shows that in that case ZF too would be inconsistent. – tomasz Jul 30 '12 at 12:05
up vote 8 down vote accepted

The continuum hypothesis can be formulated in two different ways:

  1. $2^{\aleph_0}=\aleph_1$,
  2. $A\subseteq\mathbb R$ then either $|A|\leq\aleph_0$ or $|A|=|\mathbb R|$.

It is clear that in ZFC both those statements are equivalent, but this is not the case in ZF.

However if we assume that $2^{\aleph_0}\geq\aleph_2$ then both these formulations are provably false, because it means that there is a set of real numbers which is of size $\aleph_1$ and the real numbers themselves have at least $\aleph_2$ many elements. In this case this statement is false in ZF and in ZFC.

Generally speaking, ZF and ZFC are theories, we can talk about provability; for a statement to be true we have to talk about a model. For example, AC itself is true in a model of ZFC which is a model of ZF, but it is not provable from ZF.

If a statement is provable from ZF, then it has to be provable from ZFC, we only added an axiom, so we did not make provable into unprovable or provably false. Equally, if something is provably false from ZF then it will remain provably false in ZFC.

To add a bit on the point in tomasz' answer:

It was proved by Gödel that if ZF is consistent then ZF+GCH+AC is consistent (in particular ZF+$2^{\aleph_0}=\aleph_1$ is consistent). However Cohen invented the method of forcing and used it to prove that if ZFC+GCH is consistent then ZF+$2^{\aleph_0}=\aleph_2$ is consistent.

This means that we cannot prove from ZFC alone what is the exact $\alpha$ such that $2^{\aleph_0}=\aleph_\alpha$. Namely, we cannot prove that it is $\aleph_1$ or $\aleph_2$ or higher. In fact we can make it arbitrarily high $\alpha$. Furthermore we cannot even prove from ZF that such $\alpha$ exists at all.

In order to have a theory strong enough to prove what is the exact cardinal of the continuum we need to have more axioms than merely ZFC.

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Very nice answer. Thank you – Katlus Jul 30 '12 at 14:29
more here should not be taken too literally. ZFC already has (countably) infinitely many axioms. :) – tomasz Jul 30 '12 at 15:30
@tomasz: I never knew that "more" meant only increase in cardinality! :-) – Asaf Karagila Jul 30 '12 at 16:15

I think you misunderstood Asaf Karagila's answer in that previous question.

He meant that it is possible that $\aleph_2\leq 2^{\aleph_0}$, not that it is a theorem of ZF (or ZFC), which really only means that (if ZFC is consistent) you can't prove that $\aleph_2> 2^{\aleph_0}$ in ZFC.

There can't be a statement that can be proved in ZF but not in ZFC, simply because ZFC has all the axioms of ZF and the axiom of choice. Any proof which you perform in ZF can also be done in ZFC (you need not account for axioms you don't use).

So if you could prove a statement in ZF, and prove that it is false in ZFC, that would mean that ZFC is inconsistent, and as a consequence (by Gödel's equiconsistency theorem), so is ZF, rendering much of modern mathematics quite dubious. This is not impossible, but considered unlikely by the vast majority of mathematicians.

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Now it's much clear thank you – Katlus Jul 30 '12 at 14:25

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