M
453
S
A
5
O
8, 2004
Exercise 4 from Section 18, page 111
We check that
f
is an imbedding; the argument for
g
is similar.
f
is clearly a bijection
between
X
and
f
(
X
)
=
X
×
y
0
, so we need only show that it and its inverse are continuous.
It follows from the deﬁnition of the subspace topology that the open sets of
X
×
y
0
are of
the form
U
×
y
0
where
U
is open in
X
. Since
f
(
U
)
=
U
×
y
0
, it is immediate that both
f
and
f

1
are continuous.
±
Exercise 5 from Section 18, page 111
The map
f
: [0
,
1]
→
[
a
,
b
] given by
f
(
x
)
=
a
+
(
b

a
)
x
is clearly a homeomorphism that
restricts to a homeomorphism from (0
,
1) to (
a
,
b
).
±
Exercise 6 from Section 18, page 111
Deﬁne
f
:
R
→
R
by
f
(
x
)
=
0 if
x
is rational and
f
(
x
)
=
x
otherwise. We will show that
f
is continuous only at 0 by using the sequence criterion for continuity. If
{
a
i
}
is any sequence
that converges to 0, then
{
f
(
a
i
)
}
is obtained from
{
a
i
}
by replacing the rational
a
i
’s by 0, so
it also converges to 0. Hence
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '05
 SMITH
 Math, Topology, Topological space, product topology

Click to edit the document details