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M
453
S
A
7
N
1, 2004
Exercise 3 from Section 23, page 152
Let
B
α
=
A
∪
A
α
. Then
B
α
is connected for each
α
since
A
and
A
α
are connect and have
a point in common. Now
S
B
α
=
A
∪
S
A
α
and the
B
α
’s have a point in common (
A
is
nonempty) so
A
∪
S
A
α
is connected.
±
Exercise 10 from Section 24, page 158
Suppose
x
0
∈
U
and let
V
be the set of points in
U
that can be joined to
x
0
by a path in
U
.
V
is nonempty since it contains
x
0
. If
x
∈
V
, then there is an open ball
B
(
x
, ε
) around
x
contained in
U
as
U
is open. Each point of
B
(
x
, ε
) can be connected to
x
via a straight
line (path) in
B
(
x
, ε
), and
x
can be connected to
x
0
via a path in
U
. Hence each element of
B
(
x
, ε
) can be joined to
x
0
via a path in
U
. Thus
B
(
x
, ε
)
⊂
V
and so
V
is open.
Now let
x
be a limit point of
V
in
U
. Again we can ﬁnd an open ball
B
(
x
, ε
) around
x
that is contained in
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 Spring '05
 SMITH
 Math

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