# The cardinality of the set of all finite subsets of an infinite set

Let $$X$$ be an infinite set of cardinality $$|X|$$, and let $$S$$ be the set of all finite subests of $$X$$. How can we show that Card($$S$$)$$=|X|$$? Can anyone help, please?

Addendum by LePressentiment: I have some supplementaries and recast the answer below.

$$1.$$ How and why, is that first red bracket, the upper bound on $$|S_n|$$?
$$2.$$ Whence does that equality in the 2nd red bracket originate?
$$3.$$ How would you envisage/prefigure/presage to define $$S$$ wrt $$S_n$$?
$$4.$$ If the question hadn't divulged that $$Card(S) = |X|$$, then how can $$Card(S)$$ be determined?

Since ● $$X$$ is given as infinite which means $$|X| \le |\mathbb{N}|$$
and ● $$S \quad \supseteq \quad \{\emptyset, \{1\}, \{2\},...,\{n\},...\} = \{\text{singletons}\} = |\mathbb{N}|$$ thus $$\color{#009900}{|X| \le |\mathbb{N}| \le |S|}$$.

For all $$n \in \mathbb{N}$$, define $$S_n$$ as did Prof Magidin: $$S_n:= \{\text{all subsets of cardinality }n\} \subseteq S.$$
This definition involves a subset of cardinality $$n$$ (ie with $$n$$ elements) so we now scrutinise it.
It has the form $$\{(x_1, ..., x_n)\}$$. The $$n$$ elements in this $$n$$-tuple can be chosen $$n!$$ ways, so every such subset determines $$n!$$ $$n$$-tuples of elements of $$X$$. Plainly, an $$n$$-tuple included in a set will produce a subset (of that set) of cardinality $$\le n$$. This paragraph implies: $$|S_n| \color{#B8860B}{\le} {\Large{\color{red}{[}}} \, n!|X|^n \, {\Large{{\color{red}{]}}}} = |X|$$. In toto,

\begin{align} |X| \color{#009900}{\le} |S| & = \left|\biguplus_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n| = |\underbrace{S_0}_{= \emptyset}| + \sum_{n=1}^{\infty}|S_n| \\ & = \sum_{n=1}^{\infty}|S_n| \\ & \color{#B8860B}{\le} \sum_{n=1}^{\infty}|X| = |\mathbb{N}||X| \; {\Large{\color{red}{[}}} = {\Large{{\color{red}{]}}}} \; |X| \end{align}