MATH 290 – HOMEWORK 9 – SOLUTIONS
Let
A
and
B
be sets, and suppose that
f
:
A
→
B
.
1. Prove that for every
X
⊆
A
,
X
⊆
f

1
(
f
(
X
)).
Solution:
Let
X
⊆
A
and suppose that
x
∈
X
. If we let
y
=
f
(
x
), then by deﬁnition,
y
∈
f
(
X
)
so that
f
(
x
)
∈
f
(
X
). This implies that there is an element of
X
(namely
x
) such that
f
(
x
)
∈
f
(
X
). Hence
x
∈
f

1
(
f
(
X
)).
2. Prove that
f
is injective (that is, onetoone) if and only if for every
X
⊆
A
,
X
=
f

1
(
f
(
X
)).
Solution:
(=
⇒
). Suppose that
f
is injective, and let
X
⊆
A
. Since we already know that
X
⊆
f

1
(
f
(
X
)), all we need to prove is that
f

1
(
f
(
X
))
⊆
X
. Let
x
∈
f

1
(
f
(
X
)).
Since
x
∈
f

1
(
f
(
X
)),
f
(
x
)
∈
f
(
X
) which means that there is a
y
∈
X
such that
f
(
x
) =
f
(
y
)
∈
f
(
X
). But since
f
is injective,
f
(
x
) =
f
(
y
) implies that
x
=
y
.
Therefore,
x
=
y
∈
X
.
(
⇐
=). We will prove this implication by contrapositive. Suppose that
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 Fall '08
 Alligood,K
 Advanced Math, Sets, Inverse function, Basic concepts in set theory

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