Let $X$ be an infinite set of cardinality $|X|$, and let $S$ be the set of all finite subsets of $X$. How can we show that Card($S$)$=|X|$? Can anyone help, please?
Cardinality of Finite Subsets – Infinite Set
elementary-set-theory
elementary-set-theory
Let $X$ be an infinite set of cardinality $|X|$, and let $S$ be the set of all finite subsets of $X$. How can we show that Card($S$)$=|X|$? Can anyone help, please?
Best Answer
The cardinality is at least $|X|$, since $S$ contains all singletons.
Let $S_n$ be the subset of $S$ consisting of all subsets of cardinality exactly $n$. Then $S$ is the disjoint union of the $S_n$.
Now, for any positive integer $n$, the number of subsets of $X$ of cardinality $n$ is at most $|X|^n = |X|$ (equality since $|X|$ is infinite); because an $n$-tuple of elements of $X$ determines a subset of $X$ of cardinality at most $n$; and every subset with $n$ elements determines only finitely many distinct $n$-tuples of elements of $X$ (namely, $n!$). So $|S_n|\leq n!|X|^n = |X|$ for all $n$.
Therefore: \begin{align*} |X| &\leq |S| = \left|\bigcup_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n|\\ &= \sum_{n=1}^{\infty}|S_n| \leq \sum_{n=1}^{\infty}|X|\\ &= \aleph_0|X| = |X|, \end{align*} with the last equality since $|X|\geq\aleph_0$.
Thus, $|X|\leq |S|\leq |X|$, so $|S|=|X|$.