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An Alternative Proof of Bolzano's Theorem

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Abstract

We use the completeness of the real numbers along with the definition and the sequential criterion of continuity to prove a lemma from which Bolzano's theorem follows easily.
An Alternative Proof of Bolzanos Theorem
Spiros Konstantogiannis
spiroskonstantogiannis@gmail.com
Abstract. We use the completeness of the real numbers along with the definition and the
sequential criterion of continuity to prove a lemma from which Bolzanos theorem follows easily.
Keywords: completeness of the real numbers; continuity; Bolzanos theorem.
1 Auxiliary lemma
Lemma
Let
I
Ì
¡
be a right-closed interval and let
:f I
®
¡
be continuous on
I
. If the set
(
)
: 0
S x I f x
-
= Î <
is non-empty, then
(i)
sup
S
-
exists and belongs to
I
.
(ii) If
sup
S S
- -
Î
, then
sup
S
-
is the right endpoint of
I
.
(iii) If
sup
S S
- -
Ï
, then
f
vanishes at
sup
S
-
, i.e.
(
)
sup 0
f S
-
=
.
Proof
(i) Since
I
is a right-closed interval, then it contains its right endpoint, say
b
.
By assumption,
S
-
is non-empty.
Further, if
x S
-
Î
, then, by the definition of the set
S
-
,
x I
Î
, and since
b
is the right endpoint of
I
,
x b
£
, thus
S
-
is bounded above by
b
.
Next, since
S
-
is non-empty and bounded above, then, by the completeness of
¡
,
sup
S
-
exists
in
¡
.
Also, since, by definition,
sup
S
-
is the least upper bound of
S
-
, and
b
is an upper bound of
S
-
,
then
sup
S b
-
£
(1).
Further, as
S
-
is non-empty,
S
-
has at least one element, say
0
x
, and then
0
sup
x S b
-
£ £
(2).
Besides, since
0
x S
-
Î
, then, by the definition of
S
-
,
0
x I
Î
.
Then
0
,
x b I
Î
.
If
0
x b
=
, then, by (2),
sup
S b I
-
= Î
.
If
0
x b
¹
, then, by (2),
0
x b
<
, and since
0
,
x b I
Î
and
I
is an interval, then
[
]
0
,
x b I
Í
, and by
(2),
[
]
0
sup ,
S x b I
-
Î Í
, thus
sup
S I
-
Î
.
Thus, in both cases,
sup
S I
-
Î
.
(ii) We will show by contradiction that
sup
S b
-
=
.
To this end, we assume that
sup
S S
- -
Î
and we also assume that
sup
S b
-
¹
.
Since
sup
S S
- -
Î
, then
(
)
sup 0
f S
-
<
(3).
Next, by (i),
sup
S I
-
Î
, and since
f
is continuous on
I
, then
f
is continuous at
sup
S
-
, and
thus, by the definition of continuity, for every
0
e
>
there exists
0
d
>
(that depends on both
e
and
sup
S
-
) such that
(
)
(
)
supf x f S
e
-
- <
for all
x I
Î
and supx S
d
-
- <
.
Since the last inequality is equivalently written as
{
(
)
Adding
sup to
all sides
sup sup sup sup ,sup
S
x S S x S x S S
d d d d d d
-
- - - - -
- < - < Û - < < + Û Î - +
,
we have that
(
)
(
)
supf x f S
e
-
- <
for all
(
)
sup ,sup .
x I S S
d d
- -
Î Ç - +
Since the last inequality holds for every
0
e
>
, then it also holds for
(
)
sup 0
f S
e
-
= - >
(by (3)),
and thus there exists
0
0
d
>
(remember that
d
depends on
e
) such that
(
)
(
)
(
)
sup sup
f x f S f S
- -
- < - , from which it follows that
(
)
(
)
(
)
sup sup
f x f S f S
- -
- < - ,
and then
(
)
0
f x
<
(4), for all
(
)
0 0
sup ,supx I S S
d d
- -
Î Ç - + .
Besides, by assumption,
sup
S b
-
¹
, and combining with (1) yields
sup
S b
-
<
, and then combining
with the inequality
sup supS S
d
- -
< +
yields
{
}
0
sup min ,supS b S
d
- -
< + .
Next, since the reals are dense in themselves, the last inequality yields that there exists
1
x
Î
¡
such
that
{
}
1 0
sup min ,supS x b S
d
- -
< < + (5), from which it follows that
1
sup
S x b
-
< <
(6) and
1 0
sup supS x S
d
- -
< < +
(7).
Next, from (6), we have that
[
]
1
sup ,
x S b
-
Î.
Also, by (i),
sup
S I
-
Î
, and
b I
Î
. Then, since
I
is an interval,
[
]
sup ,
S b I
-
Í
, and combining
with
[
]
1
sup ,
x S b
-
Î yields
1
x I
Î
.
Next, from (7), we have that
(
)
(
)
1 0 0 0
sup ,sup sup ,supx S S S S
d d d
- - - -
Î + Ì - + , thus
(
)
1 0 0
sup ,supx S S
d d
- -
Î - + .
Thus
1
x I
Î
and
(
)
1 0 0
sup ,supx S S
d d
- -
Î - + , from which it follows that
(
)
1 0 0
sup ,supx I S S
d d
- -
Î Ç - + , and then, by (4),
(
)
1
0
f x
<
.
Finally, since
1
x I
Î
and
(
)
1
0
f x
<
, then, by the definition of the set
S
-
,
1
x S
-
Î
, and since
sup
S
-
is an upper bound of
S
-
,
1
sup
x S
-
£
, which contradicts (5).
Therefore the assumption that
sup
S b
-
¹
is false, and then
sup
S b
-
=
, which completes the proof.
(iii) Since
sup
S
-
is the supremum of
S
-
, then for every
0
e
>
there exists
x S
-
Î
such that
supx S
e
-
> -
, and since
sup
S
-
is also an upper bound of
S
-
that, in our case, does not belong
to
S
-
, then
sup
x S
-
<
, and combining the last two inequalities yields
sup sup
S x S
e
- -
- < <
.
Then choosing
1 0
n
e
= >
, we have that for every n
Î
¥
there exists
n
x S
-
Î
such that
sup 1 sup
n
S n x S
- -
- < <
(8).
Next, since
(
)
lim 1 0
n
=
, then both sequences
(
)
sup 1
S n
-
- and
(
)
sup
S
-
converge to
sup
S
-
, and then, by the squeeze theorem for sequences, (8) yields that
(
)
n
x
also converges to
sup
S
-
, i.e.
(
)
lim sup
n
x S
-
= (9).
Besides, by the definition of the set
S
-
, if
x S
-
Î
, then
x I
Î
, and thus
S I
-
Í
.
Next, since, for every n
Î
¥
,
n
x S I
-
Î Í
, then
n
x I
Î
, and thus the sequence
(
)
n
x
is in
I
.
Further, by (i),
sup
S I
-
Î
, and since
f
is continuous on
I
, then
f
is continuous at
sup
S
-
.
Then the sequence
(
)
n
x
is in
I
, converges to
sup
S I
-
Î
, and
f
is continuous at
sup
S
-
.
Thus, by the sequential criterion for continuity, the sequence
(
)
(
)
n
f x
converges to
(
)
(
)
lim
n
f x
,
which, by (9), is equal to
(
)
sup
f S
-
, i.e.
(
)
(
)
(
)
lim sup
n
f x f S
-
= (10).
Besides, since
sup
S I
-
Î
and
sup
S S
- -
Ï
, then
(
)
sup 0
f S
-
³
, and by (10),
(
)
(
)
lim 0
n
f x
³
(11).
On the other hand, since
n
x S
-
Î
, then
(
)
0
n
f x
<
, for all n
Î
¥
.
The sequence
(
)
(
)
n
f x
has then strictly negative terms and, as we showed, it converges.
Then, by the properties of sequence limit, its limit is non-positive, i.e.
(
)
(
)
lim 0
n
f x
£
, and
combining with (11) yields
(
)
(
)
lim 0
n
f x
=
, and by (10),
(
)
sup 0
f S
-
=
, which completes the
proof.
Corollary
Let
I
Ì
¡
be a right-closed interval and let
:f I
®
¡
be continuous on
I
. If the set
(
)
: 0
S x I f x
+
= Î >
is non-empty, then
(i)
sup
S
+
exists and belongs to
I
.
(ii) If
sup
S S
+ +
Î
, then
sup
S
+
is the right endpoint of
I
.
(iii) If
sup
S S
+ +
Ï
, then
f
vanishes at
sup
S
+
, i.e.
(
)
sup 0
f S
+
=
.
Proof
Since
f
is continuous on
I
, then
f
-
is also continuous on
I
.
Also
(
)
: 0 .
S x I f x
+
= Î - <
Then applying the previous lemma yields
(i)
sup
S
+
exists and belongs to
I
.
(ii) If
sup
S S
+ +
Î
, then
sup
S
+
is the right endpoint of
I
.
(iii) If
sup
S S
+ +
Ï
, then
f
-
vanishes at
sup
S
+
, i.e.
(
)
sup 0
f S
+
- =
, thus
(
)
sup 0
f S
+
=
, i.e.
f
vanishes at
sup
S
+
.
2 Bolzanos theorem
If
[
]
: ,f a b ®
¡
is continuous on
[
]
,
a b
and
(
)
(
)
0
f a f b
<
, then there exists
(
)
,
c a b
Î such
that
(
)
0.
f c
=
Proof
Since
(
)
(
)
0
f a f b
<
, then
(
)
0
f a
<
and
(
)
0
f b
>
, or
(
)
0
f a
>
and
(
)
0
f b
<
.
(i)
(
)
0
f a
<
and
(
)
0
f b
>
.
Considering the set
[
]
(
)
{
}
, : 0
S x a b f x
-
= Î <
, we observe that
[
]
,
a a b
Î and
(
)
0
f a
<
, thus
a S
-
Î
, and then
S
-
is non-empty.
Also, the interval
[
]
,
a b
is right-closed.
Then, by the previous lemma,
[
]
sup ,
S a b
-
Î.
Further, if
sup
S S
- -
Î
, then, by the previous lemma,
sup
S b
-
=
, thus
b S
-
Î
, and then, by the
definition of the set
S
-
,
(
)
0
f b
<
, which is a contradiction.
Therefore
sup
S S
- -
Ï
, and then, again by the previous lemma,
(
)
sup 0
f S
-
=
.
Finally, since
(
)
(
)
0
f a f b
¹
, then
sup ,
S a b
-
¹
, and combining with
[
]
sup ,
S a b
-
Î yields that
(
)
sup ,
S a b
-
Î, which completes the proof.
(ii)
(
)
0
f a
>
and
(
)
0
f b
<
.
Considering the set
[
]
(
)
{
}
, : 0
S x a b f x
+
= Î >
, we observe that
[
]
,
a a b
Î and
(
)
0
f a
>
, thus
a S
+
Î
, and then
S
+
is non-empty.
Also, the interval
[
]
,
a b
is right-closed.
Then, by the previous corollary,
[
]
sup ,
S a b
+
Î.
Further, if
sup
S S
+ +
Î
, then, by the previous corollary,
sup
S b
+
=
, thus
b S
+
Î
, and then, by the
definition of the set
S
+
,
(
)
0
f b
>
, which is a contradiction.
Therefore
sup
S S
+ +
Ï
, and then, again by the previous corollary,
(
)
sup 0
f S
+
=
.
Finally, since
(
)
(
)
0
f a f b
¹
, then
sup ,
S a b
+
¹
, and combining with
[
]
sup ,
S a b
+
Î yields that
(
)
sup ,
S a b
+
Î, which completes the proof.
3 References
Robert G. Bartle and Donald R. Sherbert, Introduction to Real Analysis. John Wiley & Sons, Inc., Fourth Edition,
2011.
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ResearchGate has not been able to resolve any references for this publication.