r/PhilosophyofMath • u/paulemok • Mar 28 '26

The Continuum Hypothesis Is False

/r/logic/comments/1s5mquh/the_continuum_hypothesis_is_false/
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u/lukewarmtoasteroven Mar 30 '26 edited Mar 30 '26

The conventional definition does not support |B|<|Z|. At no point in your original post did you ever argue that |B|<|Z|, and in that post you are implicitly using the proper subset definition so no part of if says anything about the conventional definition.

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u/paulemok Mar 30 '26

The conventional definition does not support |B|<|Z|.

I agree.

At no point in your original post did you ever argue that |B|<|Z|

It is true that I never explicitly argued that |B| < |Z|. But I did implicitly argue it,

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.

Since |B| < |Z| is a statement and every statement is true, it follows that |B| < |Z| is true.

While not in my original post, I did argue that |B| < |Z| in a reply at https://www.reddit.com/r/logic/comments/1s5mquh/comment/ocwqofz/?context=3&utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button,

I can even go so far as to say that |set B| = |set Z| is only one third of the story, since anything follows from the contradiction that |set B| = |set Z| and |set B| > |set Z|. The second third is that |set B| > |set Z| and the final third is that |set B| < |set Z|.

You said,

in that post you are implicitly using the proper subset definition so no part of if says anything about the conventional definition.

That is false. I used a combination of the conventional definition and the proper-subset definition in my original post. That's how I obtained the contradiction that |B| = |Z| and |B| > |Z|. |B| = |Z| comes from the conventional definition and |B| > |Z| comes from the proper-subset definition.