# Cardinality of the Set of all finite subset of $\mathbb{R}$

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<...

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.

• @avs The power set is the set of all subsets, the OP asked for the set of all finite subsets. Dec 13, 2016 at 23:49
• 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.) Dec 14, 2016 at 0:08
• It's easier if you know that $|\mathbb R|=|\mathcal P(\mathbb N)|.$ Mar 27, 2019 at 20:18

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 $$|\mathcal{P}_{<\omega}(\mathbb{R})|=|\mathbb{R}|=2^{\aleph_0}$$