Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Was asked this question, and I have no idea about how to start proving it.

Could someone give me some good reference material to start with.

share|cite|improve this question
It is unproveable in ZFC set theory if there is a cardinal between $\aleph_0$ (rationals) and $2^{\aleph_0}$ (irrational). – William Sep 17 '12 at 6:21
This is called the continuum hypothesis. Googling will give you good results. – Alex Becker Sep 17 '12 at 6:21
Dear William: it is quite unlikey that you comment is of help to the OP... – Mariano Suárez-Alvarez Sep 17 '12 at 6:22
@MarianoSuárez-Alvarez at least it give me terms that I should look up and learn more about. Nothing is really lost in asking stupid questions, even if others answers are for now beyond my knowledge. – yiyi Sep 17 '12 at 6:23
To elaborate. The question is related to the continuum hypothesis. You can produce models of $ZFC$ where $\aleph_1 = 2^{\aleph_0}$ and a model where $\aleph_1 < 2^{\aleph_0}$. In the former, there is no cardinality in between. In the latter, there is. Hence the axioms of $ZFC$ can not prove the continuum hypothesis. – William Sep 17 '12 at 6:24
up vote 14 down vote accepted

The assertion that there is none is called The Continuum Hypothesis, and it is provably unprovable from the usual axioms of mathematics. Cantor made this hypothesis some 150 years ago, and it was in fact the first on Hilbert's problems.

Historically Cantor asked this a long time before there was a formal foundation to set theory. Some time into the 20th century the formulation of set theory as we know it today began to take shape. Now we use an axiom system called ZFC, originally formulated by Zermelo, with additional axioms by Fraenkel, and with the axiom of Choice adjoined.

Gödel proved in the subsequent years that if a theory is consistent then it has a model, and much later he showed that if ZF is consistent then ZFC+CH (and more) is consistent. He did this by constructing what is known as The Constructible Universe in modern set theory.

In 1963 Paul Cohen came up with a technique which is known as Forcing, in which we begin with a model of set theory and we extend it by adding a particular set which has certain properties. Using this technique he showed that if ZFC+CH is consistent then it is so is ZFC+"CH is false".

Combining the two results we have that from the usual axioms of set theory (namely, ZFC) we cannot prove nor disprove the continuum hypothesis. Both Gödel's work as well Cohen's are complicated and technical. Most books about axiomatic set theory should cover them at some point. For example Kunen's or Jech's.

share|cite|improve this answer
Unless we accept law of excluded middle there is no reason to regard or disregard CH. – Norbert Sep 17 '12 at 13:16
@Norbert: I don't like math where you can't prove by contradiction... – Asaf Karagila Sep 17 '12 at 13:22
So do I${}{}{}$ – Norbert Sep 17 '12 at 13:51
@Norbert so you like math where you can't prove by contradiction? – yiyi Sep 18 '12 at 7:23
@Norbert: I think that you may have missed the double negation in my previous comment; or that you may have applied some intuitionistic logic to it and did not deduce that the double negation can be eliminated... :-) – Asaf Karagila Sep 20 '12 at 22:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.