Let’s define $C_\alpha^\beta$ as the cardinality of the set of all subsets with cardinality $\beta$ of a set with cardinality $\alpha$:

$$C_\alpha^\beta = |\{T \subset S| |T| = \beta \}|$$

where $|S| = \alpha$.

It is easy to see, that $C_\alpha^\beta$ is well defined, and if $\alpha$ and $\beta$ are finite, then $C_\alpha^\beta$ becomes classical binomial coefficients.

It is also true, that $C_\alpha^\beta = 0$ for all $\beta > \alpha$.

Now one can see, that for infinite $\alpha$ and finite $\beta$ we have $C_\alpha^\beta = \alpha$.

It is also true, that for infinite $\alpha$ $C_\alpha^\alpha = 2^{\alpha}$:

$|\{T \subset S| |T| = |S| \}| \geq |\{T \subset S| |T| < |S| \}|$, as if $|T| < |S|$, then $|S \setminus T| = |S|$, for infinite $S$.

$$2^{|S|} = |\{T \subset S \}| = |\{T \subset S| |T| = |S| \}| + |\{T \subset S| |T| < |S| \}| \leq |\{T \subset S| |T| = |S| \}| \leq |\{T \subset S \}| = 2^{|S|}$$

However, I do not know the values of $C_\alpha^\beta$, where $\alpha$ and $\beta$ are both infinite and $\beta < \alpha$. Is there a way to find them too?


marked as duplicate by Asaf Karagila cardinals May 2 at 11:54

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