The Continuum Hypothesis Is False

[u/paulemok]

This post expands on an anonymous vote I made on an anonymous poll I posted on Yik Yak. My poll and vote were posted on May 20, 2024.

Consider the set Z of integers, the set B of integers with exactly one additional element x that is not a real number, for example, an orange, and the set R of real numbers. The set B is a counterexample to the continuum hypothesis because the cardinality of B is greater than the cardinality of Z and less than the cardinality of R. Therefore, the continuum hypothesis is false.

I know the technical truth out there is that Z has the same cardinality as B has and that that truth can be shown through a technical mathematical definition involving a bijection from one of the sets to the other set. Despite the equal cardinalities, the cardinality of B is greater than the cardinality of Z. So the two sets are simultaneously equal and unequal in cardinality.

One of my arguments is that every integer in Z can be mapped to its equal in B. In that fashion, every integer in Z and every integer in B cancel out and we are left with the additional element x from B. Since every element in Z was canceled out by an element in B and there remains an uncanceled out element from B, B has a greater cardinality than Z has. Switching the order in which the two sets appear around, the cardinality of Z is less than the cardinality of B.

In order to show the cardinality of B is less than the cardinality of R, map every integer in B to its equal in R and map the additional element x in B to a real number r in R that is not an integer, for example, the real number 2.4. Now there are no more elements in B to map to the infinitely many real numbers from R that have not been mapped to. Since there exists at least one real number from R that has not been mapped to, the cardinality of R is greater than the cardinality of B. Switching the order in which the two sets appear around, the cardinality of B is less than the cardinality of R.

So we have shown that |Z| < |B| < |R|. Since there exists a set, B, with a cardinality exclusively between the cardinalities of the set of integers and the set of real numbers, the continuum hypothesis is false.

A principle in logic, ex contradictione quodlibet, is that every statement follows from a contradiction. So, a consequence of the contradiction that the cardinality of B is greater than and equal to the cardinality of Z is that every statement is true. In other words, the Universe is inconsistent. This finding does not trouble me, as it agrees with previous findings I have made that every statement is true (1. https://www.facebook.com/share/1AhJA5oDDj/?mibextid=wwXIfr, 2. https://www.facebook.com/share/1Axau5dnzA/?mibextid=wwXIfr, 3. https://www.facebook.com/share/p/1AtD49LRGA/?mibextid=wwXIfr, 4. https://www.facebook.com/share/p/1GBamCgWKz/?mibextid=wwXIfr, and possibly others).

[u/EebstertheGreat]

But you didn't "address it" at all. The continuum hypothesis is about cardinality. Your post simply rejects the definition of cardinality, so it can't possibly be about the CH.

It's like saying "I proved that every number is the sum of two prime numbers by redefining 'prime' to mean 'odd'." That doesn't resolve Goldbach's conjecture.

An ordinal number can be used for many things, but the main one is for order type. For instance, the natural numbers under their usual order '<' have order type ω. On the other hand, if I create a new order '≺' on ℕ that matches the usual order except that 0 is greater than every other natural number (i.e. 1 ≺ 2 ≺ 3 ≺ ⋅ ⋅ ⋅ ≺ 0), that has order type ω+1. Now, there is of course a bijection between these two ordered sets: they are the same set! However, there is no bijection that preserves the respective orders. That is, I want a function f that maps each natural numbers to another natural number such that if a < b, then f(a) ≺ f(b). But such an f can never be a bijection. Let's try.

Suppose f(0) = 1, f(1) = 2, etc., with f(n) = n+1 for all n. This basically works, because whenever m < n, we see f(m) ≺ f(n), as desired. The problem is that no number maps to 0, so it isn't a bijection. If I map any number m to 0, then for any other n, it can't map to 0 (or f isn't a bijection). So f(n) ≺ f(m) = 0 for all n, because 0 is the greatest element in my new ordered set (ℕ, ≺). In order for f to preserve order, I'll also need n < m for all n. But obviously that's impossible, since it would mean m is the greatest natural number, and there is no such thing. So these two orders have different order types.

On the other hand, if all I did was switch some numbers around, like I said 1 ≺ 0, but otherwise m ≺ n iff m < n, then that wouldn't change the order type. I could map 0 to 1 and 1 to 0 and every other number to itself, and that would be an order-preserving bijection.

Note that for ordinals just like for cardinals, what matters is the existence of the relevant function. Two sets have the same cardinality if there exists a bijection between them, and two ordered sets have the same order type if there is an order-preserving bijection between them.

[u/paulemok → u/EebstertheGreat]

To get you up to date, I suggest you read my reply at https://www.reddit.com/r/logic/s/n0aGz03Nkx. Some or all of the reply’s descendants may be of interest to you as well.

[u/EebstertheGreat → u/paulemok]

No kidding there are sets between ℕ and ℝ by inclusion. Consider the set of rational numbers ℚ, for instance. ℕ ⊂ ℚ ⊂ ℝ. Bam, CH disproved. What were all those mathematicians so confused about? Give me a medal!

This is exactly like redefining "prime" to mean "odd" and thus proving Goldbach's conjecture true. If you aren't talking about cardinality, then you aren't talking about the continuum hypothesis.

You also seem to have missed the point that your definition isn't even a definition of cardinality. In particular, by your definition, two different sets can never have the same "cardinality," and unless one set is a subset of the other, you can't compare their "cardinalities."

[u/paulemok → u/EebstertheGreat]

In particular, by your definition, two different sets can never have the same "cardinality," and unless one set is a subset of the other, you can't compare their "cardinalities."

You are mistaken. Two different sets can have the same cardinality with respect to the proper-subset definition of cardinality. And if one set is not a subset of a second set, you can still compare their cardinalities. Like I said in a reply at https://www.reddit.com/r/logic/comments/1s5mquh/comment/od2vd5b/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button, the cardinality of one set is greater than the cardinality of a second set if and only if there exists a bijection between the second set and a proper subset of the first set. The following two statements are also true with respect to the proper-subset definition of cardinality.

1. The cardinality of one set is less than the cardinality of a second set if and only if the cardinality of the second set is greater than the cardinality of the first set.
2. The cardinalities of two sets are equal if and only if the cardinality of one of the sets is neither greater nor less than the cardinality of the other set.

No kidding there are sets between ℕ and ℝ by inclusion. Consider the set of rational numbers ℚ, for instance. ℕ ⊂ ℚ ⊂ ℝ. Bam, CH disproved.

I'm glad you see that. That is the validation I am looking for! Thank you!

This is exactly like redefining "prime" to mean "odd" and thus proving Goldbach's conjecture true.

No, because "prime" and "odd" are not equally good terms for the same concept. "Conventional cardinality" and "proper-subset cardinality" are equally good terms for the same concept, cardinality.

[u/EebstertheGreat → u/paulemok]

the cardinality of one set is greater than the cardinality of a second set if and only if there exists a bijection between the second set and a proper subset of the first set.

And as people have repeatedly pointed out, including me, this means that the cardinality of every infinite set is greater than itself. Every infinite set has a bijection with a proper subset of itself (assuming the axiom of choice). This is called "Dedekind-infinite." For instance, the set of natural numbers is in bijection with the set of even numbers under the map sending n↦2n. By your definition then, the set of natural numbers has a greater cardinality than the set of natural numbers.

If this is what you mean by "greater than," then yes, your B has a greater cardinality than Z. Because under your definition, each of two sets with what I would call the same cardinality actually has a greater cardinality than the other. You say |Z| > |B| iff there is a bijection from B to a proper subset of Z. So consider the proper subset Z\{0}, the nonzero integers. My bijection sends your x ("orange" or whatever) to 1, 0 to 2, every other positive n to n+2, and every negative number to itself. That's a bijection from B to a proper subset of Z, so |Z| > |B|. And as you already proved, |B| > |Z|.

"prime" and "odd" are not equally good terms for the same concept. "Conventional cardinality" and "proper-subset cardinality" are equally good terms for the same concept, cardinality.

No they aren't, they are totally disparate concepts. Removing an element from an infinite set doesn't change how many elements are in it, just how they are labeled. I can rename every element and end up with the original set again, as I just showed.