# "Disproving" Cantor's hypothesis -- it's trivial, anyway

## Recommended Posts

I think the set of counting integers cannot scale against a set with more members as bases, each base of which can be exponentiated to an infinite set itself. But, this is difficult to conceptualize, and I do not know what all axioms we are working under. Apparently ordered pairs are not controversial, but one exemplar to the contrary (or to the contrary of 'X is countable therefore f(X) is countable') can upend an axiom.

On 4/25/2022 at 2:47 PM, uncool said:

[...] for whatever exponent set you choose (possibly subject to some rare exceptions).

What do you hint at here? Furthermore, would you recommend any reference on set theory, or other reading/topics on number theories?

Also, how do you see a new model of the reals under forcing? In my opinion it is a fine piece of logical proof, but I read into it that it is pointing out how there are infinitely many numbers between any two points on the line because of unlimited decimal extensions.

##### Share on other sites

Again: the set of algebraic numbers is countable. There is a bijection between the set of integers and the set of algebraic numbers.

And by making an infinite set of exponents, you are again making a map from a countable set (the set of pairs of bases and exponents). The image of a countable set under any map is countable. This isn’t that hard to prove - it’s not an axiom, it’s a theorem.

Edited by uncool

##### Share on other sites

• 5 months later...
On 4/21/2022 at 7:02 AM, uncool said:

The set of all transcendentals has cardinality equal to that of the reals.

I don't think we actually know that. It may be that the complex numbers have a lot more transcendentals than the reals, don't you suppose???

On 4/21/2022 at 7:02 AM, uncool said:

The set of all square roots of primes has cardinality equal to that of the natural numbers, by rather obvious bijection.

Ok, but to reinforce: the reals have cardinality in part defined by having all transcendentals you claimed, and the naturals have none.

On 4/21/2022 at 7:02 AM, uncool said:

Neither of those have “intermediate” cardinality. If you are claiming to have constructed a set of intermediate cardinality, what is it?

The results obtained from the function $\sqrt{i'}^{\sqrt{i'}$

On 4/21/2022 at 7:02 AM, uncool said:

That we will force the natural numbers set to have to match each choice for base, against which all results from exponentiation could be counted; but we will also force a querying function: each counting number is only able to query for a digit, so each counting number can only count out decimals against a single resultant transcendental, unable to account for a single base and set of exponentiated results in my opinion. Then, moving through the change of base and subsequent exponentiations and forced querying operations the counting numbers simply cannot account properly for the set of transcendentals generated by this self-referential set operation. Too much to count.

On 4/21/2022 at 7:02 AM, uncool said:

Cardinality is not about underlying members. Just because the elements of a set are transcendentals and there are lots of transcendentals doesn’t mean that set is large.

I think the set is both infinite counting up with the naturals, and infinite counting out the decimals -- because that's what seperates naturals and reals. While you assert that the set of all transcendentals is of cardinality with the reals, I assert that this set generated is of intermediate cardinality, as in a similar way the transcendentals of the reals not included here cannot be accounted for by those that are here in the i' set.

On 4/27/2022 at 10:51 AM, uncool said:

Part of the point of “forcing” is that it is not constructing new sets within the same model - which is your approach - but instead constructing a new model of the real numbers.

My idea is to construct a new model of the complex numbers.

Thanks, @uncool, I also appreciated a post on QP you made recently, I'll try to find it to +1 it.

edit: I was only reviewing page 1, sorry... I thought I hadn't gotten to this somehow.

Edited by NTuft
clarity.
##### Share on other sites

Are you aware that The Continuum Hypothesis can neither be proved nor disproved ?

##### Share on other sites

On 10/3/2022 at 7:34 PM, NTuft said:

I don't think we actually know that. It may be that the complex numbers have a lot more transcendentals than the reals, don't you suppose???

No. They are, in fact, in bijection, and both are in bijection with the set of all real numbers, and both are in bijection with the set of all complex numbers.

On 10/3/2022 at 7:34 PM, NTuft said:

Ok, but to reinforce: the reals have cardinality in part defined by having all transcendentals you claimed, and the naturals have none.

Again: there is nothing inherently uncountable about individual transcendental numbers. You keep bringing up that some of these numbers are transcendental as if that, by itself, implies uncountability. It does not.

On 10/3/2022 at 7:34 PM, NTuft said:

The results obtained from the function $\sqrt{i'}^{\sqrt{i'}$

First: what is your argument for uncountability?

Second: that set is very clearly countable. It is very clearly the image of a countable set, namely the set of pairs (x, y) where x and y are both in i', under the map of exponentiation.

Once again: emphasizing that you are getting results which are transcendental numbers does not tell you anything about whether the output you get is uncountable. At the absolute best, it removes a single, overly-simplistic proof of countability; that does not constitute a proof of uncountability.

## Create an account

Register a new account