Problem Set 5 Solutions
Problem 1:
(The given statement is not true, so we will state and prove its negation.)
1. Statement:
There are sets
A, B, X
, with
X
⊆
A
, and a function
f
:
A
→
B
, such that
f
−
1
(
f
(
X
)) is not a subset of
X
.
2. Proof:
Let
A
=
{
a
1
, a
2
}
, X
=
{
a
1
}
, B
=
{
b
}
. Then
X
⊆
A
. Let
f
:
A
→
B
be defined
by
f
(
a
1
) =
b
and
f
(
a
2
) =
b
.
Since
a
1
∈
X
and
f
(
a
1
) =
b
,
b
∈
f
(
X
). And since
b
∈
f
(
X
) and
f
(
a
2
) =
b
,
a
2
∈
f
−
1
(
f
(
X
)).
Since
a
2
∈
f
−
1
(
f
(
X
)) but
a
2
∈
X
,
f
−
1
(
f
(
X
))
⊆
X
.
Problem 2
Lemma 1:
f
(
f
−
1
(
Y
))
⊇
Y
.
Proof of Lemma 1:
Let
b
∈
Y
. Since
f
is onto,
∃
a
∈
A
such that
f
(
a
) =
b
.
Since
f
(
a
) =
b
and
b
∈
Y
,
a
∈
f
−
1
(
Y
).
This implies that
f
(
a
)
∈
f
(
f
−
1
(
Y
)).
But since
b
=
f
(
a
), this means that
b
∈
f
(
f
−
1
(
Y
)).
Lemma 2:
f
(
f
−
1
(
Y
))
⊆
Y
.
Proof of Lemma 2:
Let
b
∈
f
(
f
−
1
(
Y
)). Then
∃
a
∈
f
−
1
(
Y
) such that
f
(
a
) =
b
.
Since
a
∈
f
−
1
(
Y
),
f
(
a
)
∈
Y
. Since
f
(
a
) =
b
, this means that
b
∈
Y
.
Claim:
f
(
f
−
1
(
Y
)) =
Y
.
Proof of Claim:
Since
f
(
f
−
1
(
Y
))
⊇
Y
(by Lemma 1) and
f
(
f
−
1
(
Y
))
⊆
Y
(by Lemma 2),
f
(
f
−
1
(
Y
)) =
Y
.
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 Spring '08
 Foster
 Assumption of Mary, Lemma

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