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Solutions to homework # 8.
1. No, this does not imply that
f
is continuous. Consider the function
f
(
x
) :=
±
1
/

x

if
x
6
= 0
,
0
x
= 0
.
Then
f
is not continuous at
x
= 0 even though
f
(
x
)

f
(

x
) = 0 for all
x
∈
IR.
2.
For every
x
∈
E
,
f
(
x
)
∈
f
(
E
)
⊆
f
(
E
), hence
x
∈
f

1
(
f
(
E
)). Thus
E
⊆
f

1
(
f
(
E
)).
The last set must be closed as the preimage of the closed set
f
(
E
) by the Corollary to
Theorem 4.8, hence it also contains
E
. So,
E
⊆
f

1
(
f
(
E
)) =
⇒
f
(
E
)
⊆
f
(
f

1
(
f
(
E
)))
⊆
f
(
E
)
.
Here is an example showing that the inclusion
f
(
E
)
⊆
f
(
E
) may be proper. Take
X
to be Q
and
Y
to be IR and let
f
be simply the identity map, identifying rational points as members
of IR. Take
E
to be Q
, too. Since
E
is all of Q
, it is closed (in Q
), but its closure in IR is
all of IR, hence
f
(
E
) = IR
6
= Q
=
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This document was uploaded on 11/17/2010.
 Spring '09

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