3.3: Power Sets
- Page ID
- 22652
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Given a set \(A,\) we call the set of all subsets of \(A\) the power set of \(A,\) which we denote \(\mathcal{P}(A)\).
If \(A=\{1,2,3\},\) then
\[\mathcal{P}(A)=\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}.\]
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.
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 \(A .\)
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 \(|A|<|B| .\)
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.
\(\mathbb{R}\) is uncountable.
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. \(|I|=|\{x: x \in \mathbb{R},-1<x<1\}|\)
Let \(I=\{x: x \in \mathbb{R}, 0 \leq x<1\}\) and suppose \(a\) and \(b\) are real numbers with \(a<b .\) Show that
\[|I|=|\{x: x \in \mathbb{R}, a \leq x<b\}|.\]
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.)