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Mathematics LibreTexts

3.3: Power Sets

  • Page ID
    22652
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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.)

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