September 29, 2008
Munkres
§
18
Ex. 18.1 (Morten Poulsen).
Recall the
ε

δ
deﬁnition of continuity: A function
f
:
R
→
R
is
said to be continuous if
∀
a
∈
R
∀
ε
∈
R
+
∃
δ
∈
R
+
∀
x
∈
R
:

x

a

< δ
⇒ 
f
(
x
)

f
(
a
)

< ε.
Let
T
be the standard topology on
R
generated by the open intervals.
Theorem 1.
For functions
f
:
R
→
R
the following are equivalent:
(
i
)
∀
a
∈
R
∀
ε
∈
R
+
∃
δ
∈
R
+
∀
x
∈
R
:

x

a

< δ
⇒ 
f
(
x
)

f
(
a
)

< ε
.
(
ii
)
∀
a
∈
R
∀
ε
∈
R
+
∃
δ
∈
R
+
:
f
((
a

δ,a
+
δ
))
⊂
(
f
(
a
)

ε,f
(
a
) +
ε
)
.
(
iii
)
∀
U
∈ T
:
f

1
(
U
)
∈ T
.
Proof.
“(
i
)
⇔
(
ii
)”: Clear.
“(
ii
)
⇒
(
iii
)”: Let
U
∈ T
. If
a
∈
f

1
(
U
) then
f
(
a
)
∈
U
. Since
U
is open there exists
ε
∈
R
+
such that (
f
(
a
)

ε,f
(
a
) +
ε
)
⊂
U
. By assumption there exists
δ
a
∈
R
+
such that
f
((
a

δ
a
,a
+
δ
a
))
⊂
(
f
(
a
)

ε,f
(
a
) +
ε
), hence (
a

δ
a
,a
+
δ
a
)
⊂
f

1
(
U
). It follows that
f

1
(
U
) =
S
a
∈
f

1
(
U
)
(
a

δ
a
,a
+
δ
a
), i.e. open.
“(
iii
)
⇒
(
ii
)”: Let
a
∈
R
. Given
ε
∈
R
+
then
f

1
((
f
(
a
)

ε,f
(
a
) +
ε
)) is open and contains
a
. Hence there exists
δ
∈
R
+
such that (
a

δ,a
+
δ
)
⊂
f

1
((
f
(
a
)

ε,f
(
a
) +
ε
)). It follows that
f
((
a

δ,a
+
δ
))
⊂
(
f
(
a
)

ε,f
(
a
) +
ε
).
±
Ex. 18.2.
Let
f
:
R
→ {
0
}
be the constant map. Then 2004 is a limit point of
R
but
f
(2004) = 0
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.
 Fall '08
 Brown
 Topology, Continuity, Metric space, Open set, R+, continuous maps, Morten Poulsen

Click to edit the document details