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Let $\kappa , \lambda$ be cardinals with $\omega \leq \lambda \leq \kappa.$

Prove $\kappa^{\lambda} = |\{X: X \subseteq \kappa, |X|=\lambda\}|$.

i.e could anyone advise me on how to construct a bijection between $\{f \mid \text{$f$ is a function},~{\rm dom}(f) = \lambda,~ {\rm ran}(f) \subseteq \kappa\}$ and $\{X: X \subseteq \kappa, |X|=\lambda\}$? Thank you.

Lord_Farin
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Alexy Vincenzo
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1 Answers1

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Given a subset $X$ of $\kappa$ of size $\lambda$, fix a bijection $f_X:\lambda\to X$. Note $f_X$ is a function from $\lambda$ to $\kappa$ and the assignment $X\mapsto f_X$ is injective.

Conversely, recalling that a function is the same as its graph, any function $f:\lambda\to\kappa$ is a subset of $\lambda\times\kappa$ of size $\lambda$. But $\lambda\times\kappa$ is in bijection with $\kappa$, so we can identify $f$ with a subset $X_f$ of $\kappa$ of size $\lambda$. This assignment is also injective.

We are now done by the Bernstein-Schroeder theorem.

Andrés E. Caicedo
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  • For the first paragraph, one can choose some function from $\lambda$ to $\kappa$ having a given range of size $\lambda$. But I don't think "increasing enumeration" works in general. For example: $\omega+\omega$ as a subset of $\omega_1$ has no increasing enumeration from $\omega$? – PatrickR Feb 28 '24 at 06:07
  • @PatrickR Yes, thanks. I was probably thinking of $[\kappa]^\lambda$ instead, which in some contexts means the set of subsets of $\kappa$ of order type $\lambda$, and ended up writing something which was a mixture of the two. Fixed now. – Andrés E. Caicedo Feb 28 '24 at 16:12