My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Reply
 
LinkBack Thread Tools Display Modes
October 11th, 2018, 01:19 PM   #11
Senior Member
 
Joined: Jun 2014
From: USA

Posts: 422
Thanks: 26

Quote:
Originally Posted by cjem View Post
Interesting. Given that the rest of the argument looks fine (and assuming ZF is consistent), this means $H$ isn't an element of $G$ for any surjection $r$. It seemed plausible to me that $H$ would at least sometimes be an element of $G$, even if I wasn't convinced it always would be.
I may be able to salvage the proof if I explicitly define function $r$ so that it is bijective, prove $H$ in $G$, and then derive the contradiction anyways. I see it in my head but won’t be able to write it up until later.
AplanisTophet is offline  
 
October 12th, 2018, 03:31 AM   #12
Senior Member
 
Joined: Jun 2014
From: USA

Posts: 422
Thanks: 26

Quote:
Originally Posted by cjem View Post
Interesting. Given that the rest of the argument looks fine (and assuming ZF is consistent), this means $H$ isn't an element of $G$ for any surjection $r$. It seemed plausible to me that $H$ would at least sometimes be an element of $G$, even if I wasn't convinced it always would be.
Yup, that's about what it boils down to, "$H$ isn't an element of $G$ for any surjection $r$."

Cantor made use of "in / not in" operations to construct his self-referential argument whereas I made use of "subset / not subset" operations to construct mine. If I had made $G$ larger so as to contain the set $H$ for any possible function $r$, then the cardinality of $G$ would be greater than the cardinality of $D$. Cantor's proof that no set can be surjected onto its powerset was obviously much more eloquent, but this is another way of showing the difference in cardinality between two sets. So, if anyone doesn't like Cantor's arguments (it seems we have a few running around this forum), maybe they could accept my argument as constructed with a suitable $G$.
AplanisTophet is offline  
October 12th, 2018, 04:42 AM   #13
Senior Member
 
Joined: Aug 2017
From: United Kingdom

Posts: 284
Thanks: 86

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
Originally Posted by AplanisTophet View Post
So, if anyone doesn't like Cantor's arguments (it seems we have a few running around this forum), maybe they could accept my argument as constructed with a suitable $G$.
Unfortunately, at least one of these people think they have proven $\mathbb N$ is in bijection with the unit interval. So, to them, any argument to the contrary is automatically wrong and therefore isn't even worth engaging with. They seem to have a very weird and self-contradictory intuition for sets of strings: they seem to insist that the set of all finite strings (note: they don't reject the existence of such a set) on some alphabet must contain infinite strings.

This is one reason that learning to write proofs and establishing rigour early on in your mathematical education is important: it gives you the ability to challenge any potentially flawed intuitions before they root themselves too deeply in your mind.
cjem is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
opus



Thread Tools
Display Modes






Copyright © 2018 My Math Forum. All rights reserved.