Find the cardinality of the set of all finite subsets of $\mathbb{R}$.

I have proved that the set of all finite subsets of $\mathbb{N}$ is countable . But I cannot find the cardinality of the set in case of $\mathbb{R}$ .

My Attempt:

First I have considered the set $$A_k=\{\{a_1, a_2, ..., a_k\}| a_i \in \mathbb{R} \ and\ a_1<...<a_k \}$$

Then $S=$ The set of all finite subsets of $\mathbb{R}=\bigcup_{n=1}^{\infty}A_{n}$,
and now in the case of $\mathbb{N}$, I could show that this $A_k$ is countable and consequently the set of all finite subset of $\mathbb{N}$ is countable.

But in this case I cannot say anything like this...In this case the set will be of the cardinality as that of $\mathbb{R}$ (I think). But cannot prove it.

I appreciate your help. Thank you.

  • 3
    $\begingroup$ @avs The power set is the set of all subsets, the OP asked for the set of all finite subsets. $\endgroup$
    – David
    Dec 13, 2016 at 23:49
  • 2
    $\begingroup$ Can you prove that the countable union of sets, each of which has the same cardinality as $\Bbb R$, also has the cardinality of $\Bbb R$? (This is related to multiplication of cardinals, if you know anything about that at this point.) $\endgroup$ Dec 14, 2016 at 0:08
  • $\begingroup$ It's easier if you know that $|\mathbb R|=|\mathcal P(\mathbb N)|.$ $\endgroup$ Mar 27, 2019 at 20:18

2 Answers 2


We can show that $\#S \ge {c}$ via the injection $f: \mathbb{R} \rightarrow S$, $f(x) = \{x\}$.

If you show $\#S \le c$ then you can conclude $\#S = c$ invoking Bernstein's theorem.

Let's prove the following: let $X = \{X_k,\ k \in \mathbb{N}\}$ be a family of subsets such that $\#X_i \le c$, $\ \forall i \in \mathbb{N}$, and let $V = \bigcup_{k\in \mathbb{N}}X_k$. Then $\#V \le c$.

Because $\#X_i \le c$, we have for each $i$ a surjective $f_i: \mathbb{R} \rightarrow X_i$. We define $g: \mathbb{N} \times \mathbb{R} \rightarrow V$, $g(n,x) = f_n(x)$.

$g$ is an surjective function: let $x \in V$, then $x \in X_i$ for some $i$, then we have $a \in \mathbb{R}$ such that $f_i(a) = x$ because $f_i$ is surjective. Then $g(i,a) = f_i(a) = x$, with $(i,a) \in \mathbb{N}\times\mathbb{R}$.

We conclude that $g$ is surjective.

Thus $\#V \le \#(\mathbb{N}\times\mathbb{R})$, this is, $\#V \le c$.

Your set $S$ satisfies the hypotheses, so we have $\#S \le c$, and thus $\#S = c$.


Let $A$ be an element of the set of all the finite subsets of $\mathbb{R}$, which we shall denote simply by $\mathcal{P}_{<\omega}(\mathbb{R})$

Assertion: $\mathcal{P}_{<\omega}(\mathbb{R})\preccurlyeq\,^\omega\mathbb{R}$

We will identify $A$ with a $\omega$-sequence of elements of $\mathbb{R}$, that is, and element of $^\omega\mathbb{R}$, in the following way:

Assume that $A=\{a_0,\dots,a_n\}$. Then we can construct a $\omega$-sequence $(b_k)_{k\in\omega}$ defined by:

$$b_k=\begin{cases} a_k\qquad\qquad\text{if }k\le n \\ a_n+1\qquad\text{ if }k>n \end{cases}$$

It is clear that the correspondence $A\longmapsto(b_k)_{k\in\omega}$ is injective.

Now, on the one hand we have that $\mathbb{R}\preccurlyeq\mathcal{P}_{<\omega}(\mathbb{R})$, because the function $r\in\mathbb{R}\longmapsto\{r\}\in\mathcal{P}_{<\omega}(\mathbb{R})$ is obviously injective.

On the other hand, $\mathcal{P}_{<\omega}(\mathbb{R})\preccurlyeq\mathbb{R}$, since $\;\mathcal{P}_{<\omega}(\mathbb{R})\preccurlyeq\,^\omega\mathbb{R}\;$ and $\;^\omega\mathbb{R}\preccurlyeq\mathbb{R}$: in fact, $|^\omega\mathbb{R}|=\big(2^{\aleph_0}\big)^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}=|\mathbb{R}|$

From the Cantor-Bernstein therem, we obtain that $|\mathcal{P}_{<\omega}(\mathbb{R})|=|\mathbb{R}|=2^{\aleph_0}$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .