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Suppose
f
is a function with domain [
a,b
]. Since we will be concerend with
subsets of [
a,b
], we will simplify notation by letting (
p,q
)
·
stand for (
p,q
)
∩
[
a,b
].
If
X
is a subset of [
a,b
],
f
(
X
) :=
{
f
(
x
)

x
∈
X
}
.
Suppose
c
∈
[
a,b
]. Recall what we mean when we say that
f
is continuous at
c
:
For any real number
± >
0, there is a real number
δ >
0 so that:
for all
x
∈
[
a,b
],

x

c

< δ
implies

f
(
x
)

f
(
c
)

< ±
.
The condition above (oﬀset on its own line) is equivalent to:
for all
x
∈
(
c

δ,c
+
δ
)
·
,
f
(
x
)
∈
(
f
(
c
)

±,f
(
c
) +
±
)
,
and to:
f
(
(
c

δ,c
+
δ
)
·
)
⊆
(
f
(
c
)

±,f
(
c
) +
±
)
.
Fact.
If
X
is a bounded subset of the real line and
u
is the least upper bound
of
X
, then any open interval that contains
u
also contains points of
X
.
Proof.
If
u
were surrounded by an open interval containing no ponts of
X
, there would
be points in that interval to the left of
u
that were upper bounds for
X
. Then
u
could not be the least upper bount.
Boundedness Theorem.
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This note was uploaded on 11/23/2011 for the course MATH 1551 taught by Professor Malisoff during the Fall '08 term at LSU.
 Fall '08
 MALISOFF
 Calculus, Sets

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