July 13th, 2017, 06:49 PM  #11  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522  Quote:
What makes you think it's a set at all? I'm pretty sure it isn't. Do you mean order isomorphisms to itself (that's a set), or to other sets (that isn't)? You've lost me here. Last edited by Maschke; July 13th, 2017 at 06:52 PM.  
July 13th, 2017, 08:20 PM  #12 
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  Order isomorphisms to itself. For example, if $\alpha = 2 = \{0,1\}$, then $\alpha^I = \{ \{0_1,1_2\}, \{1_1,0_2\}\}$. "...[E]ach isomorphism also has order type $\alpha$ and every element of $\alpha$ appears in each isomorphism."
Last edited by AplanisTophet; July 13th, 2017 at 08:25 PM. 
July 14th, 2017, 08:40 AM  #13 
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522  
July 14th, 2017, 08:46 AM  #14 
Senior Member Joined: Oct 2009 Posts: 402 Thanks: 139 
I'm sorry, I find this hard to follow. What is an order isomorphism from $\omega$ to $\omega$? For me, this is a bijection $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $x<y$ iff $f(x)<f(y)$. Isn't it clear that there is only one such bijection since $\omega$ has a canonical order? So I am unsure what is meant here. 
July 14th, 2017, 08:48 AM  #15 
Senior Member Joined: Oct 2009 Posts: 402 Thanks: 139 
On second thought, the natural analogue of $\aleph_0!$ would be the number of bijections from $\omega$ to $\omega$, is this what you want?

July 14th, 2017, 12:31 PM  #16 
Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26  I was playing with orderings trying to see if there was some divergence between the hierarchy of cardinalities for possible orderings of a given set and other functions on the set, like taking its powerset. Similarly, I can make the following conjecture: There is no set $A$ such that the powerset of $A$ has cardinality equal to $\aleph_0$. If $A \in \mathbb{N}$, then $P( \,A) \, \in \mathbb{N}$. If $A = \mathbb{N}$, then $P( \,A) \, > \mathbb{N}$. 
July 14th, 2017, 04:41 PM  #17  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522  Quote:
But then again as a heuristic, consider the finite case. Between $2^n$ and $2^{n+1}$ many cardinals get skipped. This is in fact one of the heuristics for the falsity of the Continuum Hypothesis. Except for cardinalities $0$ and $1$, the Peano successor of a finite number is never the same as the cardinality of the power set of a number, so perhaps that rule continues for transfinite cardinalities. Power sets are a far more powerful operation than taking successors. That argument is mentioned by Cohen.  

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