number of finite subsets of a cardinal $\kappa$

Question

Why there are only $\kappa$ many finite subsets of a cardinal $\kappa$? There are obviously at least $\kappa$ of them, but why also at most $\kappa$ of them?

Answer 1

You can show that the set $[\kappa]^n$ of $n$-subsets of $\kappa$ has cardinality $\kappa$; that is, $|[\kappa]^n|=\kappa$. (Put your mouse on the box below if you have no idea how to prove this.)

It uses the Cantor-Bernstein theorem. You can easily see that there are at least $\kappa$ $n$-subsets of $\kappa$, for example you can consider $\{0,1,\cdots, n-2, \alpha\}$ for $n-1\le \alpha < \kappa$. That is, we have $|[\kappa]^n|\ge \kappa$. Conversely, there is an obvious surjection from $\kappa^n$, a set of all sequences over $\kappa$ of length $n$, to $[\kappa]^n$. Therefore, by considering the right inverse of this onto function, which is one-to-one, we have $|[\kappa]^n|\le \kappa^n = \kappa$.

We know that the set $[\kappa]^{<\omega}$ of all finite subsets of $\kappa$ is $\bigcup_{n\in\mathbb{N}} [\kappa]^n$. Since $[\kappa]^{<\omega}$ has at least $\kappa$ elements (namely $\{\alpha\}$, for each $\alpha<\kappa$) we have $|[\kappa]^{<\omega}|\ge \kappa$. On the other hand, $$|[\kappa]^{<\omega}|= \left|\bigcup_{n\in\mathbb{N}} [\kappa]^n \right| \le \sum_n |[\kappa]^n|\le \sum_n\kappa = \kappa\cdot\aleph_0 = \kappa$$

so $|[\kappa]^{<\omega}| = \kappa$.

Answer 2

Loosely Speaking: there are (at most) $\kappa^n$ many subsets of size $n$ of a cardinal $\kappa$. Then the number of finite subsets of $\kappa$ is $$\sum_{n \in \mathbb{N}} \kappa^n = \sum_{n \in \mathbb{N}} \kappa = \aleph_0 \cdot \kappa = \kappa .$$

(Note that we are assuming $\kappa$ is infinite.)