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453
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16, 2004
Exercise 4 from Section 20, page 127
(a) Since each of their component functions are continuous, all three of
f
,
g
and
h
are
continuous when
R
ω
is given the product topology.
On the other hand, if
R
ω
is given the box topology, none of them is continuous. To
see this, look at the open set
U
=
Q
∞
n
=
1
(

1
/
n
2
,
1
/
n
2
) in
R
ω
. The inverse image of
U
under each of
f
,
g
and
h
is
{
0
}
and hence is not open.
Things are a little more interesting in the uniform topology:
f
is not continuous,
since
f

1
(
V
)
=
{
0
}
if
V
is the product of countably many copies of (

1
,
1). Let
k
denote either
g
or
h
. Then ¯
ρ
(
k
(
s
)

k
(
t
))
=
¯
d
(
s

t
), so if we have an
ε
ball
B
¯
ρ
(
f
(
s
)
, ε
)
about
f
(
s
) in the ¯
ρ
metric, the
ε
ball around
s
(i.e. the interval (
s

ε,
s
+
ε
)) in
R
will be mapped into
B
¯
ρ
(
f
(
s
)
, ε
). Hence both
g
and
h
are continuous in the uniform
topology.
(b) First observe that if (
a
n
) denotes any of the given sequences, then
π
i
(
a
n
)
→
0 for each
i
. It follows from Exercise 18.6 (and a similar proof for the uniform topology) that
the only point the sequences can possibly converge to is
0
=
(0
,
0
, . . .
). Furthermore,
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 Spring '05
 SMITH
 Topology, Topological space, uniform topology

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