$|[x]^{\lt \kappa}| = |x|^{\lt \kappa} ?$

Asked Apr 24 '23 at 00:59

Active Apr 24 '23 at 04:32

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I want to know whether $|[x]^{\lt \kappa}| = |x|^{\lt \kappa}$ holds or not.

I know $|[x]^{\lt \kappa}| \leq |x|^{\lt \kappa}$ holds. The reason is following:
There exist a order $\leq$ so that $\langle x, \leq \rangle$ become well-ordered.
For each $y \in [x]^{\lt \kappa}$, there is only one order isomorphism $f_y$ : type($x, \leq$) $\rightarrow x$.
$f_y \in {^{\lt \kappa}}x$. and $F : [x]^{\lt \kappa} \rightarrow {^{\lt \kappa}}x, y \mapsto f_y$ is injection.

However I don't know $\geq$ holds or not. Because $G : {^{\lt \kappa}}x \rightarrow [x]^{\lt \kappa}, f \mapsto \text{range}(f)$ is not injection necessarily.

Notice:
System of this site alerted to me that this question might be duplicate of already existing question $|[\kappa]^\lambda| = \kappa^\lambda$.
But that is wrong. Difference is whether '$\lt$' exist as exponential or not.

edited Apr 24 '23 at 04:32

asked Apr 24 '23 at 00:59

set_theory_jp

Notice: $\kappa^{<\lambda}=\bigcup{\kappa^\mu\mid\mu<\lambda}$. So $|\kappa^{<\lambda}|=\lambda\cdot|\sup{|\kappa^\mu|\mid\mu<\lambda}$. So, it is in fact a duplicate, with just one more small step. – Asaf Karagila Apr 24 '23 at 08:54
could you tell me more detailer proof? – set_theory_jp Apr 24 '23 at 11:43

$|[x]^{\lt \kappa}| = |x|^{\lt \kappa} ?$

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