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Umm ... Can someone disprove my proof that there are aleph-1 number of real numbers? Even comments to make my proof more rigorous are welcome.

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closed as too localized by Charles, Andrés E. Caicedo, Rudy the Reindeer, Thomas, tomasz Oct 6 '12 at 11:55

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The statement $n(B)=\aleph_1$ is the continuum hypothesis, and is not provable from the axioms of ZFC. Also, the function you define in terms of decimal expansion is not well-defined, since $.1000\cdots=.0111\cdots$. – Alex Becker Oct 4 '12 at 16:21
It seems like you are trying to show that there are as much real numbers as sets of natural numbers, Since nonen of these sets is provably equal to $\aleph_1$, this is of no help in settling the question. – Michael Greinecker Oct 4 '12 at 16:38
One usually defines $\aleph_1$ as the Hartogs number ( of $\aleph_0$, but your argument does not seem to be using this definition, so it might help to resolve your confusion if you tell us what definition of $\aleph_1$ you are using. – Trevor Wilson Oct 4 '12 at 20:26
Are you perhaps confusing the definition of $\aleph_n$ with the definition of $\beth_n$? – Trevor Wilson Oct 4 '12 at 20:33
@Trevor: While it is a reasonable (and common) mistake, not many people outside of set theory know what are $\beth$ numbers actually. Some people in set theory don't know what the $\gimel$ function is, and if we make a jump to the end of the Hebrew alphabet then most people don't know what Tav (no LaTeX symbol) is, either. – Asaf Karagila Oct 5 '12 at 0:07

This has two parts.

The first part is that assuming the axiom of choice holds, one can prove that there is a set of $\aleph_1$ many real numbers. This is not the same as saying that the real numbers themselves have size $\aleph_1$. This assertion itself is unprovable from the usual axioms of set theory and known as the Continuum Hypothesis.

On the other hand, if one is willing to drop the axiom of choice then it is conistent that there are no subset of the real numbers which has size $\aleph_1$. Namely, not only the real numbers are not of size $\aleph_1$, but there is no subset of them which has this property. It follows that the real numbers cannot be well-ordered in such model, which makes them quite peculiar as a set.

However, regardless to the assumptions about the exact size of the continuum or the axiom of choice, Cantor's theorem (which is provable without the axiom of choice) tells us that the real numbers are always uncountable.

As for the "proof" linked, there is no place where the assertion $n(B)=\aleph_1$ is explained. Instead this is just taken as an assumption, which makes the question circular -- you assume the conclusion you wish to derive.

What is true is that if $|\mathbb N\times(0,1)|=|\mathbb R|$.

For further reading:

  1. Why is the Continuum Hypothesis (not) true?
  2. Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
  3. Cardinality of sets of subsets of $\mathbb{N}$
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A one-to-one mapping (the 1st part of my proof) has been achieved between numbers between 1 and 0 and all possible subset of natural numbers. This implies they have same size. And since n(A)=n(B)=ℵ1 – Anant Saxena Oct 4 '12 at 18:41
In the foreword, I have shown how n(B)=ℵ1. Please tell me if the mapping done is wrong. Thanks – Anant Saxena Oct 4 '12 at 18:48
@Anant: Why is that set of size $\aleph_1$? Recall that the definition of $\aleph_1$ is not the cardinality of the power set of the power set of $\mathbb N$. – Asaf Karagila Oct 4 '12 at 18:48
The definition of ℵ1 being used is the size of infinity of all possible subsets of the set of natural numbers. – Anant Saxena Oct 5 '12 at 7:15
@Anant: Read that page carefully. The definition of $\aleph_1$ is the cardinality of the set of countable ordinals. The section below that of $\aleph_1$ is about the continuum hypothesis where the hypothesis is explained. – Asaf Karagila Oct 5 '12 at 9:40

There are a few "correctable" errors that aren't really important, but there is only one fatal error: you are working from the unjustified assumption that $n([0,1]) = \aleph_1$.

Since you are writing this in the first place, clearly you know it is not known (in fact not decidable) whether $n(\mathbb{R}) = \aleph_1$. But it appears you don't realize it is also not known whether $n([0,1]) = \aleph_1$ or whether $n(\mathcal{P}(\mathbb{N})) = \aleph_1$. In fact, as you have rediscovered in two different ways, these three sets are all the same cardinality, so the questions are equivalent.

In general, if you want to prove a specific set has cardinality $\aleph_1$, you have your work cut out for you. The most generally-referenced set that definitely has cardinality $\aleph_1$ is the set of countable ordinal numbers (though this is kind of a cop-out since by most definitions that set is $\aleph_1$). In fact, offhand I can't think of a way to construct a set with cardinality $\aleph_1$, without referring back to $\aleph_1$ as part of the construction. So probably the only proof strategy available is to demonstrate existence of a well-order on your set with order type $\omega_1$.

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