3.3: Power Sets
( \newcommand{\kernel}{\mathrm{null}\,}\)
Given a set A, we call the set of all subsets of A the power set of A, which we denote P(A).
If A={1,2,3}, then
P(A)={∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.
If A is finite with |A|=n, then |P(A)|=2n.
- Proof
-
Let
B={(b1,b2,…,bn):bi=0 or bi=1,i=1,2,…,n}
and let a1,a2,…,an be the elements of A. Define φ:B→P(A) by letting
φ(b1,b2,…,bn)={ai:bi=1,i=1,2,…,n}.
Then φ is a one-to-one correspondence. The result now follows from the next exercise. Q.E.D.
With B as in the previous proposition, show that |B|=2n.
In analogy with the case when A is finite, we let 2|A|=|P(A)| for any nonempty set A.
Suppose A and B are sets for which there exists a one-to-one function φ:A→ˉB but there does not exist a one-to-one correspondence ψ:A→B. Then we write |A|<|B|.
If A is a nonempty set, then |A|<|P(A)|.
- Proof
-
Define φ:A→P(Z+) by
φ({ai}∞i=1)={i:i∈Z+,ai=1}.
Then φ is a one-to-one correspondence. Q.E.D.
Now let B be the set of all sequences {ai}∞i=1 in A such that for every integer N there exists an integer n>N such that an=0. Let C=A∖B,
D0={{ai}∞i=1:ai=1,i=1,2,3,…},
and
Dj={{ai}∞i=1:aj=0,ak=1 for k>j}
for j=1,2,3,… Then |D0|=1 and |Dj|=2j−1 for j=1,2,3,… Moreover,
C=∞⋃j=0Dj,
so C is countable. Since A=B∪C, and A is uncountable, it follows that B is uncountable. Now if we let
I={x:x∈R,0≤x<1},
we have seen that the function φ:B→I defined by
φ({ai}∞i=1)=a1a2a3a4…
is a one-to-one correspondence. It follows that I is uncountable. As a consequence, we have the following result.
R is uncountable.
Let I={x:x∈R,0≤x<1}. Show that
a. |I|=|{x:x∈R,0≤x≤1}|
b. |I|=|{x:x∈R,0<x<1}|
c. |I|=|{x:x∈R,0<x<2}|
d. |I|=|{x:x∈R,−1<x<1}|
Let I={x:x∈R,0≤x<1} and suppose a and b are real numbers with a<b. Show that
|I|=|{x:x∈R,a≤x<b}|.
Does there exist a set A⊂R for which ℵ0<|A|<2ℵ0? (Before working too long on this problem, you may wish to read about Cantor's continum hypothesis.)