# Solution to the Continuum Hypothesis

Status
Not open for further replies.

#### AplanisTophet

I believe that I have solved the Continuum Hypothesis:

https://en.wikipedia.org/wiki/Continuum_hypothesis

Assume CH is true. Then $|\mathcal{P}(\mathbb{N})| = |\omega_1|$ and we have a way to prove that my $T$ sequence here cannot contain all of the elements of $\omega_1$ using the non-transfinite option:

https://math.stackexchange.com/questions/3432667/enumeration-of-a-very-large-ordinal

Alternatively, make no assumption as to whether or not CH is true. I then show above that $\omega_1$ can be placed into an $\omega$-type ordering, contradicting the definition of $\omega_1$.
$\square$

#### Micrm@ss

The Continuum Hypothesis is proven to be unprovable by the work of GÃ¶del and Cohen. I invite you to read their work and tell us where they made an error.

#### AplanisTophet

The Continuum Hypothesis is proven to be unprovable by the work of GÃ¶del and Cohen. I invite you to read their work and tell us where they made an error.
That's a lovely invitation, but there's nothing wrong with their work so I'm not sure why you would invite anyone to explain why it's wrong. If their proofs fail then it would have to be due their assumptions, such as the axioms of ZF turning out to be inconsistent or something.

I did make you look though, along with a bunch of bots apparently... In my answer on SE I say that CH is independent of my work, so yeah, I agree with you in a sense... You just assumed that what I did could be done in ZFC, but I'm using more than just those axioms unless you covert all of my ordered doublets of ordinals, triplets, etc., to sets. They do not follow the same operations as sets, so you would first have to show that you could do that before you go making any ridiculous implications that I'm intentionally trying to contradict Godel or Cohen.

I posted my answer on SE to encourage somebody to take a little interest like I normally do in my crankish, horrible, and insane fashion. That answer lacks one possibility as to $C$ though. Care to think about it?

I bet you that you can't prove $T$ doesn't equal $\omega_1$ after $\omega$ iterations using the non-transfinite option. I would think it would be easy for a good mathematician like you.

#### Micrm@ss

I bet you that you can't prove $T$ doesn't equal $\omega_1$ after $\omega$ iterations using the non-transfinite option.
Indeed, I probably can't so I won't bother trying.

In any case, if you claimed to have solved the continuum hypothesis in another axiom system, then the first thing to do is to present the axioms you used into settling it. If it is not ZFC, what is it?

#### AplanisTophet

In any case, if you claimed to have solved the continuum hypothesis in another axiom system
I just said that I have conflicting assertions here and at SE, where SE says my work is independent of CH.

present the axioms you used into settling it. If it is not ZFC, what is it?
I did (to quote the paper):

"Where $a,b$ are ordinals:

1) $a=b$ does not imply $(a,b) = (a) = (b)$.
2) $a \neq b$ does not imply $(a,b) = (b,a)$."

Indeed, I probably can't so I won't bother trying.
Then what are you posting for if other than to invite me to try and assert GÃ¶del and Cohen are wrong?

#### AplanisTophet

Let $A$ be the class of all ordinals within a model of ZFC.

Let $B$ be the class of all $t(\alpha)$ as defined in my theory where $\alpha \in Ord$ and $a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots \in Ord$.

Let an element $\alpha$ of $A$ be considered "in" an element $\beta$ of $B$ if a well ordering of all elements $x,y,z,\dots$ of $\alpha$ appears somewhere within the sequence $(a,b,c,\dots) = \beta$.

Let $f$ be a function from $A$ onto $B$.

Define the element $\kappa$ of $B$ that is the sequence of all elements $\alpha \in A$ such that they are not elements of $f(\alpha)$.

Let $c$ be such that $c \in A$ and $f(c) = \kappa$.

$c \in \kappa \implies c \not\in f(c) = \kappa$
$c \not\in \kappa \implies c \in f(c) = \kappa$

There cannot exist a $c$ where $f(c) = \kappa$ implies the class B is strictly 'larger' than the class A.

That said, GÃ¶del and Cohen working in ZFC may not be comparable to what I'm doing here. Also, I'm not sure if anyone has ever proved the size of one proper class to be larger than the size of another, but perhaps I just did... I never know.

#### greg1313

Forum Staff
We're not interested in crank posts, especially from those who are consistently rude to other members.

User banned; thread closed.

If anyone has any objections please PM us.

Last edited:
3 people
Status
Not open for further replies.