Graphics Reference
In-Depth Information
Definition.
A function f :
X
Æ
Y
is said to satisfy a
Lipschitz condition
on
X
if there
is constant M > 0, so that |f(
q
) - f(
p
)|£M |
p
-
q
| for all
p
,
q
Œ
X
.
Definition.
A function f :
X
Æ
Y
is said to be a
homeomorphism
if it is one-to-one
and onto and both it
and
its inverse are continuous.
To a topologist all homeomorphic spaces look alike and the main problem is to
classify spaces up to homeomorphism.
Definition.
A set
X
is said to be
connected
if
X
cannot be written as the union of
two subsets
A
and
B
that are nonempty disjoint open sets in
X
.
4.2.9. Theorem.
Every connected subset of
R
is either
R
itself or an interval of the
form [a,b], (a,b), (a,b], [a,b), (-•,a], (-•,a), [a,•), or (a,•).
Proof.
See [Eise74].
A more intuitive notion of connectedness is the following:
Definition.
A set
X
is said to be
path-connected
if for all points
p
and
q
in
X
there
is a continuous function f : [0,1] Æ
X
with f(0) =
p
and f(1) =
q
.
4.2.10. Proposition.
A path-connected space is connected.
Proof.
See [Eise74].
The converse of Proposition 4.2.10 is not necessarily true.
Definition.
A
component
of a set is a maximal connected subset.
Next, we describe some properties that are preserved by continuous maps.
4.2.11. Theorem.
The continuous image of a compact set is compact.
Proof.
Let
X
be a compact set and let f :
X
Æ
Y
be a continuous onto map. Let {
O
¢
a
}
be an open cover of
Y
and let
O
a
= f
-1
(
O
¢
a
). By continuity of f, each
O
a
is open in
X
.
In fact, {
O
a
} is an open cover of
X
. Since
X
is compact, there is a finite subcover
{
O
a
i
}
1£i£k
. Clearly, {
O
¢
a
i
} is a finite cover of
Y
and the theorem is proved. See also
Theorem 5.5.8.
4.2.12. Theorem.
The continuous image of a connected set is connected.
Proof.
Let
X
be a connected set and let f :
X
Æ
Y
be a continuous map which is onto
Y
. Suppose that
Y
is not connected. Then
Y
=
O
1
»
O
2
, where
O
i
is open in
Y
and
O
1
«
O
2
= f. But then
X
would be a disjoint union of open sets f
-1
(
O
1
) and f
-1
(
O
2
),
which would contradict the connectedness of
X
. Note that since f is onto, neither
f
-1
(
O
1
) nor f
-1
(
O
2
) is empty.
