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July 13th, 2017, 06:49 PM   #11
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Quote:
 Originally Posted by AplanisTophet There is no ordinal $\alpha$ such that the set of all order isomorphisms of $\alpha$ has cardinality equal to $\aleph_0$

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
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Quote:
 Originally Posted by Maschke 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.
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
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Quote:
 Originally Posted by AplanisTophet Order isomorphisms to itself.
Oh ok. Remind me what was the point of all this.

 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 $x  July 14th, 2017, 08:48 AM #15 Senior Member Joined: Oct 2009 Posts: 402 Thanks: 139 On second thought, the natural analogue of$\aleph_0!$would be the number of bijections from$\omega$to$\omega$, is this what you want? July 14th, 2017, 12:31 PM #16 Senior Member Joined: Jun 2014 From: USA Posts: 363 Thanks: 26 Quote:  Originally Posted by Maschke Oh ok. Remind me what was the point of all this. I was playing with orderings trying to see if there was some divergence between the hierarchy of cardinalities for possible orderings of a given set and other functions on the set, like taking its powerset. Similarly, I can make the following conjecture: There is no set$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}|$. July 14th, 2017, 04:41 PM #17 Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 522 Quote:  Originally Posted by AplanisTophet I can make the following conjecture: There is no set$A$such that the powerset of$A$has cardinality equal to$\aleph_0$. This is easy to prove. If$X$is finite then so is$\mathscr P(X)$and if$X$is countable then$\mathscr P(X)$is uncountable. 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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