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# 3.3: Power Sets

• • Dan Sloughter
• Professor (Mathematics) at Furman University
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## Definition

Given a set $$A,$$ we call the set of all subsets of $$A$$ the power set of $$A,$$ which we denote $$\mathcal{P}(A)$$.

## Example $$\PageIndex{1}$$

If $$A=\{1,2,3\},$$ then

$\mathcal{P}(A)=\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}.$

## Proposition $$\PageIndex{1}$$

If $$A$$ is finite with $$|A|=n,$$ then $$|\mathcal{P}(A)|=2^{n}$$.

Proof

Let

$B=\left\{\left(b_{1}, b_{2}, \ldots, b_{n}\right): b_{i}=0 \text { or } b_{i}=1, i=1,2, \ldots, n\right\}$

and let $$a_{1}, a_{2}, \ldots, a_{n}$$ be the elements of $$A .$$ Define $$\varphi: B \rightarrow \mathcal{P}(A)$$ by letting

$\varphi\left(b_{1}, b_{2}, \ldots, b_{n}\right)=\left\{a_{i}: b_{i}=1, i=1,2, \ldots, n\right\}.$

Then $$\varphi$$ is a one-to-one correspondence. The result now follows from the next exercise. $$\quad$$ Q.E.D.

## Exercise $$\PageIndex{1}$$

With $$B$$ as in the previous proposition, show that $$|B|=2^{n}$$.

In analogy with the case when $$A$$ is finite, we let $$2^{|A|}=|\mathcal{P}(A)|$$ for any nonempty set $$A .$$

## Definition

Suppose $$A$$ and $$B$$ are sets for which there exists a one-to-one function $$\varphi: A \rightarrow \bar{B}$$ but there does not exist a one-to-one correspondence $$\psi: A \rightarrow B .$$ Then we write $$|A|<|B| .$$

## Theorem $$\PageIndex{2}$$

If $$A$$ is a nonempty set, then $$|A|<|\mathcal{P}(A)|$$.

Proof

Define $$\varphi: A \rightarrow \mathcal{P}\left(\mathbb{Z}^{+}\right)$$ by

$\varphi\left(\left\{a_{i}\right\}_{i=1}^{\infty}\right)=\left\{i: i \in \mathbb{Z}^{+}, a_{i}=1\right\}.$

Then $$\varphi$$ is a one-to-one correspondence. $$\quad$$ Q.E.D.

Now let $$B$$ be the set of all sequences $$\left\{a_{i}\right\}_{i=1}^{\infty}$$ in $$A$$ such that for every integer $$N$$ there exists an integer $$n>N$$ such that $$a_{n}=0 .$$ Let $$C=A \backslash B$$,

$D_{0}=\left\{\left\{a_{i}\right\}_{i=1}^{\infty}: a_{i}=1, i=1,2,3, \ldots\right\},$

and

$D_{j}=\left\{\left\{a_{i}\right\}_{i=1}^{\infty}: a_{j}=0, a_{k}=1 \text { for } k>j\right\}$

for $$j=1,2,3, \ldots$$ Then $$\left|D_{0}\right|=1$$ and $$\left|D_{j}\right|=2^{j-1}$$ for $$j=1,2,3, \ldots$$ Moreover,

$C=\bigcup_{j=0}^{\infty} D_{j},$

so $$C$$ is countable. Since $$A=B \cup C,$$ and $$A$$ is uncountable, it follows that $$B$$ is uncountable. Now if we let

$I=\{x: x \in \mathbb{R}, 0 \leq x<1\},$

we have seen that the function $$\varphi: B \rightarrow I$$ defined by

$\varphi\left(\left\{a_{i}\right\}_{i=1}^{\infty}\right)=a_{1} a_{2} a_{3} a_{4} \ldots$

is a one-to-one correspondence. It follows that $$I$$ is uncountable. As a consequence, we have the following result.

## Theorem $$\PageIndex{4}$$

$$\mathbb{R}$$ is uncountable.

## Exercise $$\PageIndex{2}$$

Let $$I=\{x: x \in \mathbb{R}, 0 \leq x<1\} .$$ Show that

a. $$|I|=|\{x: x \in \mathbb{R}, 0 \leq x \leq 1\}|$$

b. $$|I|=|\{x: x \in \mathbb{R}, 0<x<1\}|$$

c. $$|I|=|\{x: x \in \mathbb{R}, 0<x<2\}|$$

d. $$|I|=|\{x: x \in \mathbb{R},-1<x<1\}|$$

## Exercise $$\PageIndex{3}$$

Let $$I=\{x: x \in \mathbb{R}, 0 \leq x<1\}$$ and suppose $$a$$ and $$b$$ are real numbers with $$a<b .$$ Show that

$|I|=|\{x: x \in \mathbb{R}, a \leq x<b\}|.$

## Exercise $$\PageIndex{4}$$

Does there exist a set $$A \subset \mathbb{R}$$ for which $$\aleph_{0}<|A|<2^{\aleph_{0}} ?$$ (Before working too long on this problem, you may wish to read about Cantor's continum hypothesis.)