Exam 3 Solutions
1. Using the bilinearity of the inner product we have
k
x
+
y
k
2
 k
x

y
k
2
=
h
x
+
y, x
+
y
i  h
x

y, x

y
i
=
h
x, x
i
+ 2
h
x, y
i
+
h
y, y
i 
(
h
x, x
i
+ 2
h
x,

y
i
+
h
y,

y
i
)
=
h
x, x
i
+ 2
h
x, y
i
+
h
y, y
i 
(
h
x, x
i 
2
h
x, y
i
+
h
y, y
i
) = 4
h
x, y
i
.
2. For (
x, y
)
6
= (0
,
0) we have
f
= (
π
1
·
π
2
)
·
(
g
◦
(
π
2
1
+
π
2
·
π
2
+
π
2
2
)
,
where
g
(
x
) =
1
x
. Since
x
2
+
xy
+
y
2
= (
x
+
y
)
2

xy
, it follows that (
π
2
1
+
π
2
·
π
2
+
π
2
2
)(
x, y
)
6
= 0
when (
x, y
)
6
= (0
,
0) and thus
f
is continuous at all such points. However,
f
is not continuous
at (0
,
0) because
f
(0
,
1
k
)
→
0 and
f
(
1
k
,
1
k
)
→
1
3
as
k
→ ∞
.
3. (a) Take
d
∈
f
(
S
). Then there is
c
∈
S
such that
f
(
c
) =
d
. Since
c
∈
S
there is (
x
k
) in
S
such that
x
k
→
c
. Since
f
(
x
k
)
→
f
(
c
) =
d
by the continuity of
f
and
f
(
x
k
)
∈
f
(
S
) for each
k
≥
1, we have
d
=
f
(
c
)
∈
f
(
S
).
(b) Let
f
:
R
→
R
be given by
f
(
x
) =
1
x
2
+1
. Let
S
=
Z
+
. Then
S
=
S
so
f
(
S
) =
f
(
S
) =
1
k
2
+ 1
:
k
∈
Z
+
.
Since 0
/
∈
f
(
S
) and 0
∈
f
(
S
), it follows that
f
(
S
)
6
=
f
(
S
).
4.
C
is compact so
C
is closed and since
A
is closed it follows that
A
∩
C
is closed.
Since
A
∩
C
⊆
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 Fall '09
 WYNNE
 Topology, Metric space, xk

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