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Is there a set with cardinality greater than the natural numbers but less than the real numbers?

Is there a simple proof which shows this, if the answer is no?

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Look up the continuum hypothesis. The short answer is, it's independent of the usual axioms of set theory. –  Adrian Petrescu Mar 25 '11 at 6:04
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You can also take a look at this question in MathOverflow that has some very interesting answers: mathoverflow.net/questions/23829/… –  Apostolos Mar 25 '11 at 7:03

1 Answer 1

up vote 9 down vote accepted

Yes and no. This may seem strange, so let me provide a little explanation.

There is no such thing as a proof without assumptions, so mathematicians are forced to make certain assumptions, which they take to be true without proving them, called axioms. Over the years a certain set of axioms has come into general usage among mathematicians, and most results in mathematics are proven from these axioms.

However, not every question is settled by these axioms. For your question, one cannot prove (meaning that it's been proven that no proof is possible, not just that no proof has been found yet) using the generally accepted axioms that such a set exists. However, one also cannot prove (same meaning) that no such set exists. It's kind of up to you then whether to assume the existence of such a set as a new axiom. This axiom is called the continuum hypothesis.

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I like your answer, but you meant to say the axiom is called the "Axiom of choice"! –  JavaMan Mar 25 '11 at 6:14
    
@DJC: No, the axiom of choice is an entirely separate axiom, dealing with the existence of choice functions on arbitrary sets. It is in fact one of the axioms assumed by most mathematicians. The existence of a set with cardinality between $\mathbb{N}$ and $\mathbb{R}$ is indeed referred to as "the continuum hypothesis". –  Alex Becker Mar 25 '11 at 6:17
    
@Jason: Thanks, I took your suggestion. –  Alex Becker Mar 25 '11 at 6:19
    
Interesting. Thanks. I stand corrected! –  JavaMan Mar 25 '11 at 18:12

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