hw1sol - HOMEWORK 1 SOLUTIONS 1 By induction on n Base...

HOMEWORK 1 - SOLUTIONS
1. By induction on n :
• Base Case ( n = 1): f
n
= f
1
= f which is surjective by assumption.
• Inductive Step : Suppose f n is surjective for all 1 ≤ n ≤ k . We want to show that
f k +1 is surjective. Let x ∈ X be arbitrary. We want to find y ∈ X such that
f k +1 ( y ) = x . Well f k +1 ( y ) = f ( f k ( y )) and we know f is surjective, so first pick
some z ∈ X such that f ( z ) = x . Next, by our inductive hypothesis, we know
that f
k
is surjective so pick some y such that f
k
( y ) = z . This choice of y does it,
since:
f
k +1
( y ) = f ( f
k
( y )) = f ( z ) = x
2. (GRADED - EASY). By induction on | X | :
• Base Case ( n = 0): If | X | = 0, then it’s the empty set and has only one subset,
namely itself, the empty set. And indeed:
|P ( X ) | = |{∅}| = 1 = 2
0
= 2
| X |
• Inductive Step : Assume that for all n ≤ k and every set X such that | X | = n ,
|P ( X ) | = 2 | X | . Now suppose X is a set with | X | = k + 1. Let’s write X =
{ x
1
, . . . , x
k +1
} . Now every subset of X either contains x
k +1
or it doesn’t, and it’s
easy to see that there are an equal number of subsets of X which contain x
k +1
as there are which don’t. Thus the number of subsets of X is double the number
of subsets of { x
1
, . . . , x
k
} . This latter set has cardinality k , so we can apply the
inductive hypothesis to obtain:
|P ( X ) | = 2 · |P ( { x
1
, . . . , x
k
} ) | = 2 · 2
k
= 2
k +1
3. The formula is z - 1 =
z
*
| z | 2
. To check that this works, we need to check that