On the Continuum Hypothesis

Question

Let me start out by saying that I am not a mathematician. I read an article over at Scientific American that discussed the Continuum Hypothesis. I developed the following thought experiment that would seem to prove the theory true, but as far as I know it is unprovable. Thus here I am asking for your help.

1. Start with the set of all integers $(-\infty,...,-1,0,1,...,\infty)$
2. This set can be considered 1-dimensional in that for each position in the set, there is only one possible number
3. Consider adding an additional dimension to the set that has only one possible value (0): $(<-\infty;\ 0>,...,<-1;\ 0>, <0;\ 0>, <1;\ 0>,...,<\infty; 0>)$
4. Clearly this set has the same cardinality as the integers as there is a clear mapping between them
5. Now consider having two values instead of one (0 and 5), and let's represent everything in the second dimension a decimal whose value is added to the first number
6. This makes the set $(<-\infty;\ 0;\ 5>,...,<-1;\ 0;\ 5>, <0;\ 0;\ 5>, <1;\ 0;\ 5>,...,<\infty; 0;\ 5>)$, which can be expanded to $(-\infty,...,-1.0, -0.5, 0.0, 0.5, 1.0, 1.5,...,\infty)$
7. This new set still has the same cardinality as the set of all integers since each number can be mapped directly to an integer
8. This property of the set's cardinality should continue as more numbers are added to the set, up until the point that infinitely many numbers are added
9. Once there are infinitely number of numbers added to the second dimension (sequentially), then the mapping to the integers cannot be performed because the infinite size and the maximal density (all integers from 0 to $\infty$)
10. This final set would be expanded to be exactly the same as the real numbers, and would seem to indicate that it is impossible to have a cardinality between the integers and real numbers

I guess it comes down to statement #9. It seems to me that it's most important that this set of numbers includes every combination of numbers from 0 to $\infty$ to ensure that the irrational numbers would be included.

Answer 1

The set you are building is very specific. It indeed has the same size as the reals. There are natural ways in which we can identify it with certain sets of reals. These sets are "simple", in a technical sense. It is an early theorem in descriptive set theory that all these simple sets either are countable or have the same size as the reals.

(In technical terms, one can naturally identify your set with either the irrationals, via continued fractions, or a perfect set, depending on the specific details of the construction. Even if we modify your construction, the sets you obtain this way are Borel sets. The perfect set property of a class of sets means that every set in the class is either countable or contains a perfect set. The Borel sets have the perfect set property.)

So, the fact that we jumped from countable size to the size of the reals is not due to there not being intermediate cardinalities, but instead it is a consequence of the "topological complexity" of the sets your construction produces.

One can exhibit uncountable sets of reals without perfect subsets, but all such sets are in a sense less "explicit" and instead their existence is a consequence of the axiom of choice. In certain models of set theory (that nowadays we should perhaps consider pathological), this makes them very concrete, but in other models they are less "tangible". In any case, whether any set of reals ends up having the same size as the reals or not is more subtle than the perfect set property (and, as you noted, ends up being unprovable from the standard axioms).

That said, you are in good company: Cantor proved that perfect sets have the size of the reals, and essentially started a research program whose goal was to establish the continuum hypothesis by showing that all uncountable sets of reals are "nice" in the sense that they contain perfect subsets. Now we know this is not the case, but modern research has shown that under natural assumptions (large cardinals) all "concrete" sets of reals are nice, and so, all counterexamples to the continuum hypothesis, if any, are highly non-constructive.

Answer 2

I'm not sure I understand your notation, but I think your argument boils down to the following:

1. $\mathbb{Z}$ is countable.
2. $\mathbb{Z}^2$ is countable.
3. $\mathbb{Z}^3$ is countable...
4. ...
5. $\mathbb{Z}^{\mathbb{N}}$ (the set of all countably infinite sequences of integers) has the same cardinality as $\mathbb{R}$.
6. Therefore, every subset of $\mathbb{R}$ is either countable or has the same cardinality as $\mathbb{R}$.

The last step is invalid. One thing that should immediately smell fishy about this argument is that it purports to prove something about the cardinality of an arbitrary set of real numbers, but at no point does it actually consider an arbitrary set of real numbers.

You have defined some sequence of cardinal numbers that goes"$\aleph_0, \aleph_0, \aleph_0,\ldots, 2^{\aleph_0}$, but you have no reason to believe that the cardinal numbers occurring in this sequence are the only cardinal numbers in existence. Perhaps the sequence skips over the cardinalities of some sets that you have not considered.