Math 4200
Theorem. If A is a finite set containing exactly elements, then the power set of A8
, P(A) , contains exactly 2 elements.
8
Proof:
We will prove this by mathematical induction. For each , let S( ) be the88
statement "If a set contains exactly elements, then its power set contains exactly 288
elements."
Since the only subsets of singleton set are the empty set and the set itself, the
statement S(1) is true.
We must show that for each n, S( ) implies S( +1) Suppose S( ) is true. Let88Þ8
A = be a set containing +1 elements. Any subset of A eitherÖ+ß+ßÞÞÞß+ ß+ × 8
"# 8 8+1
contains or it doesn't. The number of subsets of A which do not contain a is+8 8+1 +1
exactly the same as the number of subsets of . If S( ) is true, thisÖ+ ß + ß ÞÞÞ ß + × 8
"# 8
number is 2 . A subset of A containing a is obtained by forming the union of a
88+1
subset of with the set . Thus, the number of subsets of AÖ+ ß + ß ÞÞÞ ß + × Ö+ ×
"# 8 8+1
containing a is again 2 . The total number of subsets of A is then equal to
88
+1
2 . Hence S( +1) is true whenever S( ) is true.
88 8 8
# œ #Ð# Ñ œ # 8 8
+1
By the principle of mathematical induction, S( ) is true for all .88