MATH 202A HOMEWORK 8
Problem 1
.(a)
Proof.
Suppose
K
is compact in
α
∈
I
X
α
. Since
X
α
is closed for all
α
,
α
∈
I
X
α
is
also Hausdorff and thus
K
is closed. We just need to show that
K
◦
=
∅
.
Let
x
= (
x
α
)
∈
K
. By hint,
π
α
is continuous,
π
α
(
K
)
⊂
X
α
is compact. Since the
infinite number of
X
α
spaces are noncompact,
π
α
(
K
) =
X
α
for infinitely
α
∈
I
.
Suppose
K
◦
=
∅
, then
∃
U
=
∅
base element s.t.
U
∈
K
◦
.
U
=
α
U
α
, where
U
α
open,
U
α
=
X
α
for all but finite number
α
.
Hence,
π
α
(
U
) =
X
α
for all but finite
number
α
. However, since
U
⊂
K
,
π
α
(
U
)
⊂
π
α
(
K
), which contradicts to the above
paragraph.
(b)
Solution.
l
2
is a counterexample. Notice the identity map from
l
2
with discrete topol
ogy to
l
2
with usual
l
2
topology. By hint, any discrete topology is locally compact
since every point has a single point (itself) open neighborhood, closure of which is
compact. This map is continuous.
(c)
Proof.
Let
y
∈
Y
.
Since
f
maps from
X
onto
Y
,
∃
x
∈
X
s.t.
f
(
x
) =
y
.
Since
X
is locally compact,
∃
U
, open, containing
x
and
¯
U
is compact.
Since
f
is open,
y
=
f
(
x
)
∈
f
(
U
) is open. Hence,
f
(
U
) is a neighborhood of
y
.
We claim that
f
(
U
) is compact. We just need to show that
f
(
U
) =
f
(
¯
U
) and since
f
is continuous,
¯
U
is compact,
f
(
¯
U
) is compact.
First, we show that
f
(
U
)
⊂
f
(
¯
U
). Since
¯
U
is compact and
f
is continuous,
f
(
¯
U
)
is compact. Since
Y
is Hausdorff,
f
(
¯
U
) is closed. It is clear that
f
(
U
)
⊂
f
(
¯
U
) and
thus by definition,
f
(
U
)
⊂
f
(
¯
U
).
On the other hand, we show that
f
(
¯
U
)
⊂
f
(
U
).
Since
f
(
U
) is closed and
f
is an open map,
f

1
(
f
(
U
)) is closed in
X
.
It is clear that
U
⊂
f

1
(
f
(
U
)) and
since
f

1
(
f
(
U
)) is closed, by definition, we know that
¯
U
⊂
f

1
(
f
(
U
)).
Therefore,
f
(
¯
U
)
⊂
f
(
f

1
(
f
(
U
)))
⊂
f
(
U
).
(d)
Proof.
By (c), the projection
π
α
is continuous, surjective and open and thus we have
the conclusion.
(e)
1
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MATH 202A HOMEWORK 8
Proof.
By (a), since if infinite number
X
α
are compact, then compact sets in the
product are nowhere dense. Since
¯
U
◦
=
U
,
¯
U
cannot be compact unless
U
=
∅
.
(f)
Proof.
Let (
x
α
)
∈
α
∈
I
X
α
.
Let the index of finitely many not compact
X
α
be
α
1
, α
2
,
· · ·
, α
n
.
∀
j
∈
[1
, n
], let
U
α
j
be open sets in
X
α
j
s.t.
x
α
j
∈
U
α
j
,
U
α
j
compact
since
X
α
j
is locally compact. Let
U
=
α
∈
I
U
α
, where
U
α
=
U
α
j
when
α
=
α
j
and
U
α
=
X
α
otherwise. By Tychonoff Theorem,
¯
U
=
α
∈
I
U
α
is compact.
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 Spring '10
 tao/analysis
 Math, Topology, Metric space, Open set, General topology, vy, Xα

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