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July 13th, 2017, 06:49 PM   #11
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Originally Posted by AplanisTophet View Post
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.
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July 13th, 2017, 08:20 PM   #12
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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."

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July 14th, 2017, 08:40 AM   #13
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Order isomorphisms to itself.
Oh ok. Remind me what was the point of all this.
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July 14th, 2017, 08:46 AM   #14
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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.
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July 14th, 2017, 08:48 AM   #15
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On second thought, the natural analogue of $\aleph_0!$ would be the number of bijections from $\omega$ to $\omega$, is this what you want?
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July 14th, 2017, 12:31 PM   #16
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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 $|P( \,A) \,| > |\mathbb{N}|$.
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July 14th, 2017, 04:41 PM   #17
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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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