# Homework Help: Zakon Vol 1, Ch2, Sec-6, Prob-19 : Cardinality of union of 2 sets

1. Mar 5, 2012

### dawoodvora

1. The problem statement, all variables and given/known data

Show by induction that if the fi nite sets A and B have m and n elements,
respectively, then
(i) A X B has mn elements;
(ii) A has 2m subsets;
(iii) If further A $\cap$ B = $\varphi$, then A $\cup$ B has m+ n elements.

NOTE : I am only interested in the (iii) section of the problem. Section (i) and (ii), I was able to solve albeit with some help.

2. Relevant equations

3. The attempt at a solution
Here is the attempt at the solution

I am using mathematical induction to solve the query.

P(n) : |A| = m, |B| = n; |A $\cup$ B| = m + n if A $\cap$ B = $\varphi$.

Basis Step:

P(n = 1) to be proven as true.

Let A = { a1, a2,......am} => |A| = m
Let B = {b1} => |B| = 1

Now A $\cup$ B = {a1, a2,...am,b1}
|A $\cup$ B| = m + 1 elements, since A and B are disjoint(given) Thus P(1) holds true.
Basis Step is proven.

Let us redefine B = {b1, b2, ....., bk, bk+1} => |B| = k+1.
Also |A $\cap$ B| = $\varphi$
Now consider Bk = {b1, b2,..., bk} => Bk $\subset$ B, |Bk| = k

Inductive hypothesis:
Assume P(k) holds true for n = k i.e.,

|A $\cup$ Bk| = m + k, when A $\cap$ Bk = $\varphi$

Inductive Step:

Bk $\subset$ B => B = Bk $\cup$ {bk+1}
A $\cup$ B = A $\cup$ (Bk $\cup$ {bk+1}) = (A $\cup$ Bk) $\cup$ {Bk+1} = m + k + 1 = m + (k + 1) => P(k+1) holds true. Thus Induction is complete.

Somehow, I am not convinced with the solution that I have attempted. Please let me know if I am missing something.

2. Mar 6, 2012

### HallsofIvy

This is wrong. What you have written is 2 times m while, in fact, A has 2m subsets. If you don't want to use LaTeX (as you have below) use 2^m.

That's correct but seems overly complicated. If A and B have no members in common, $A\cup B$ contains every member of A and B- thus has m+ n members.