MATH 74 HOMEWORK 6 SOLUTIONS
1(a).
By #5 from last week’s homework,
A
n
has 2
n
subsets.
1(b).
Let
S
n
denote the set of all subsets of
A
n
with an even number of elements,
and
T
n
the set of all subsets of
A
n
with an odd number of elements.
We claim that #(
T
n
) = #(
S
n
). Do do this, we will define a function
f
:
T
n
→
S
n
and prove it is a bijection. Define
f
by
f
(
X
) =
X
∪ {
1
}
1
∈
X
X
\ {
1
}
1
∈
X
Note that no matter what, as
f
(
X
) has either one more or one less element than
X
does,
f
really does send elements of
T
n
to elements of
S
n
.
Let
g
:
S
n
→
T
n
be given by the same rule,
g
(
Y
) =
Y
∪ {
1
}
1
∈
Y
Y
\ {
1
}
1
∈
Y
For the same reason,
g
really does send elements of
S
n
to
T
n
.
Remark.
This is not the only way to do this exercise. There are a ton of other
bijections between
S
n
and
T
n
you could write down. [If the
f
and
g
I chose above
seem mysterious, write out what it they are doing in the case
n
= 3
or
4
. The idea
should be clearer then.]
Alternatively you could prove the result by induction (to
prove the inductive step, you’d need to establish some relationship between
#
S
n
+1
and
#
S
n
). There are tons of ways to do this, I only chose one of them. The same
remark applies to all of the other exercises.
For any
X
∈
T
n
, if 1
∈
X
we have
g
(
f
(
X
)) =
g
(
X
\ {
1
}
) = (
X
\ {
1
}
)
∪ {
1
}
=
X
and if 1
∈
X
we have
g
(
f
(
X
)) =
g
(
X
∪ {
1
}
) = (
X
∪ {
1
}
)
\ {
1
}
=
X
so that
g
(
f
(
X
)) =
X
for all
X
∈
T
n
. Similarly, for any
Y
∈
S
n
f
(
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 Fall '07
 COURTNEY
 Sets, Tn, Sn, consecutive pair

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