June 28th, 2013, 10:37 AM  #1 
Joined: Nov 2011 Posts: 599 Thanks: 0  Question
Let D = { 0.1011..., 0.110111..., 0.1110111..., ... } Let E = { 0.111..., 0.1011..., 0.110111..., 0.1110111..., ... } Both sets have a cardinality equal to aleph null. If E is well ordered, then does it have a last element? Does well ordering E change its cardinality? 
June 28th, 2013, 10:43 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 14,335 Thanks: 516 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Question Quote:
Quote:
 
June 28th, 2013, 10:44 AM  #3 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
0.111... would be greater than any element of D. If E is well ordered, then D is a subset of E (D is a subset of E regardless). 0.111... would be the last element of E, right?

June 28th, 2013, 10:55 AM  #4 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
Here is what you are not realizing. D = E 
June 28th, 2013, 10:58 AM  #5 
Senior Member Joined: Aug 2012 Posts: 436 Thanks: 29  Re: Question
I better get in quick before this thread gets locked. @krausebj0, I was amused by your thread over at Physicsforum. They not only locked your thread, they deleted it entirely. That's pretty rare even for the moderators over at that place. I don't even post there anymore because their moderators are so quick to lock threads they don't like. But getting your entire thread actually deleted as if it never existed ... that's pretty impressivel I did find it interesting that you got one poster to read your stuff in detail, and he came up with the same concern I did ... namely, that your assignment of a sequence of "random" reals to subsets of D is not well defined and does not yield a bijection. In fact by considering the infinite binary tree of zeros and ones I convinced myself that you can never biject the reals to the dyadics that way. In other words if you take a real .0101001100101... and assign it to the subset of its "go to the next 1" initial sequences .01, .0101, ... you can never get a bijection that way. It's true that since D is countable, you can biject it to any countable set of reals. However the "initial segment" correspondence can never be a bijection, as you can see from contemplating the infinite binary tree. I think there's a diagonal argument in there but I didn't work out all the details. So if you take a countable set of reals and biject it to D, that bijection can never be compatible with the "initial segment" correspondence you are considering. You really should carefully reconsider your steps 3 and 4, because your process is not welldefined and in fact can never result in a bijection. That's the heart of the failure of your argument. 
June 28th, 2013, 11:09 AM  #6 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
Ai is bounded. If Ai is not bounded, then Ai = 0.10111..., 0.110111..., 0.1110111..., is an infinite sequence. If Ai is bounded, then the above is not an infinite sequence. Ai is only infinitely large if A = D. 
June 28th, 2013, 11:10 AM  #7 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
If you wish to discuss my previous paper here, I can make a surjection from D onto RI just by making sure A = D. Ai is bounded.

June 28th, 2013, 11:11 AM  #8 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
Back to the point. Can anyone prove that E does not equal D? 
June 28th, 2013, 11:13 AM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 14,335 Thanks: 516 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: Question Quote:
 
June 28th, 2013, 11:14 AM  #10 
Joined: Nov 2011 Posts: 599 Thanks: 0  Re: Question
If nobody can prove that E does not equal D, then I assert E equals D. Once I assert that, you guys better stop locking my threads.
