Proofs of “Three Hard Theorems”

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Oct 8, 2013 (3 years and 8 months ago)

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Proofs of\Three Hard Theorems"
Fall 2004
Chapter x7 of Spivak's Calculus focuses on three of the most important
theorems in Calculus.In this note I will give alternative proofs of these
results.1 Preliminaries
We review a few important facts we have seen about sequences.
Lemma 1 If x
n
2 [a;b] for all n 2 N and (x
n
)
1n=1
converges to x.Then
x 2 [a;b].
Proof We ¯rst show a · x.Suppose x < a.Choose ² > 0 with ² < a ¡x.
Then no element of the sequence is in the interval (x¡²;x+²),a contradicton.
A similar argument shows b ¸ x.
Lemma 2 If f:[c;d]!R is continuous,a
n
2 [c;d] for n 2 N and (a
n
)
1n=1
converges to a 2 [c;d].Then (f(a
n
))
1n=1
converges to f(a).
Proof Let ² > 0.Since f is continuous,there is ± > 0 such that if jx¡aj < ±,
then jf(x) ¡ f(a)j < ².Since (a
n
)
1n=1
!a,there is N 2 N such that
ja
n
¡ aj < ± for all n ¸ N.Thus jf(a
n
) ¡ f(a)j < ² for all n ¸ N and
(f(a
n
))
1n=1
!f(a):
Theorem 3 (Nested Interval Theorem) Suppose I
n
= [a
n
;b
n
] where
a
n
< b
n
for n 2 N and I
1
¶ I
2
¶ I
3
¶:::.Then
1
\
n=1
I
n
6=;:
1
Proof Note that we have
a
1
· a
2
· a
3
:::· a
n
·::::::· b
n
·:::b
2
· b
1
:
Then each b
i
is an upper bound for the set A = fa
1
;a
2
;:::g.By the Com-
pleteness Axiom,we can ¯nd ® a least upper bound for A.
We claim that ® 2 I
n
for all n 2 N.Fix n 2 N.Since ® is an upper
bound for A,a
n
· ®.But b
n
is an upper bound for A and ® is the least
upper bound.Thus ® · b
n
.Hence ® 2 I
n
for all n 2 N and ® 2
1
\
n=1
I
n
.
Theorem 4 (Bolzano{Weierstrass Theorem) Every bounded sequence has
a convergent subsequence.
Proof Let (x
i
)
1i=1
be bounded.There is M 2 R such that jx
i
j · M for all
i 2 N.We inductively construct a sequence of intervals
I
0
¾ I
1
¾ I
2
¾:::
such that:
i) I
n
is a closed interval [a
n
;b
n
] where b
n
¡a
n
=
2M
2
n
;
ii) fi:x
i
2 I
n
g is in¯nite.
We let I
0
= [¡M;M].This closed interval has length 2M and x
i
2 I
0
for
all i 2 N.
Suppose we have I
n
= [a
n
;b
n
] satisfying i) and ii).Let c
n
be the midpoint
a
n
+b
n
2
.Each of the intervals [a
n
;c
n
] and [c
n
;b
n
] is half the length of I
n
.Thus
they both have length
1
2
2M
2
n
=
2M
2
n+1
If x
i
2 I
n
,then x
i
2 [a
n
;c
n
] or x
i
2 [c
n
;b
n
],
possibly both.Thus at least one of the sets
fi:x
i
2 [a
n
;c
n
]g and fi:x
i
2 [c
n
;b
n
]g
is in¯nite.If the ¯rst is in¯nite,we let a
n+1
= a
n
and b
n+1
= c
n
.If the
second is in¯nite,we let a
n+1
= c
n
and b
n+1
= b
n
.Let I
n+1
= [a
n+1
;b
n+1
]
Then i) and ii) are satis¯ed.
By the Nested Interval Theorem,there is ® 2
1
\
n=1
I
n
.We next ¯nd a
subsequence converging to ®.
2
Choose i
1
2 N such that x
i
1
2 I
1
.Suppose we have i
n
.We know
that fi:x
i
2 I
n+1
g is in¯nite.Thus we can choose i
n+1
> i
n
such that
x
i
n+1
2 I
n+1
.This allows us to construct a sequence of natural numbers
i
1
< i
2
< i
3
<:::
where i
n
2 I
n
for all n 2 N.
We ¯nish the proof by showing that the subsequence (x
in
)
1n=1
!®.Let
² > 0.Choose N such that ² >
2M
2
N
.Suppose n ¸ N.Then x
i
n
2 I
n
and
® 2 I
n
.Thus
jx
i
n
¡®j ·
2M
2
n
·
2M
2
N
< ²
for all n ¸ N and (x
i
n
)
1n=1
!®.
2 Bounding and the Extreme Value Theorem
Theorem 5 (Bounding Theorem) If f:[a;b]!R is continous,then
there is M 2 R such that jf(x)j · M for all x 2 [a;b].
Proof For purposes of contradiction,suppose not.Then for any n 2 N we
can ¯nd x
n
2 [a;b] such that jf(x
n
)j > n.By the Bolzano{Weierstrass The-
orem,we can ¯nd a convergent subsequence x
i
1
;x
i
2
;:::.Note that jf(x
i
n
j >
i
n
¸ n.Thus,replacing (x
n
)
1n=1
by (x
i
n
)
1n=1
,we may,without loss of gener-
ality,assume that (x
n
)
1n=1
is convergent.Suppose (x
n
)
1n=1
!x.By Lemma
1,x 2 [a;b].By Lemma 2,
(f(x
n
))
1n=1
!f(x):
But the sequence (f(x
n
))
1n=1
is unbounded,and hence divergent,a contradi-
cation.Theorem 6 (Extreme Value Theorem) Suppose a < b.If
f:[a;b]!R,then there are c;d 2 [a;b] such that f(c) · f(x) · f(d)
for all x 2 [a;b].
Proof Let A = ff(x):a · x · bg.Then A 6=;and,by the Bounding
Theorem,A is bounded above and below.Let ® = supA.We claim that
there is d 2 [a;b] with f(d) = ®.
3
Since ® = supA,for each n 2 N,there is x
n
2 [a;b] with
® ¡
1
n
< f(x
n
) · ®.Note that (f(x
n
))
1n=1
converges to ®.By the Bolzano{
Weierstrass Theorem,we can ¯nd a convergent subseqence.Replacing (x
n
)
1n=1
by a subsequence if necessary,we may assume (x
n
)
1n=1
!d for some d 2 [a;b].
Then (f(x
n
))
1n=1
!f(d).Thus f(d) = ®.Note that f(x) · ® = f(d) for all
x 2 [a;b].
Similarly,we can ¯nd c 2 [a;b] with f(c) = ¯ = inf A and f(c) · f(x)
for all x 2 [a;b].
3 Intermediate Value Theorem
Theorem 7 (Intermediate Value Theorem) If f:[a;b]!R is contin-
uous and f(a) < 0 < f(b),then there is a < c < b with f(c) = 0.
Proof We start to build a sequence of intervals
I
0
¶ I
1
¶ I
2
¶:::
such that I
n
= [a
n
;b
n
],f(a
n
) < 0 < f(b
n
) and b
n
¡ a
n
=
b¡a
=2
n
.Let a
0
=
a;b
0
= b and I
0
= [a
0
;b
0
].Then f(a
0
) < 0 < b
0
and b ¡a = (b ¡a)=2
0
.
Suppose we are given I
n
= [a
n
;b
n
] with f(a
n
) < 0 < f(b
n
) and b
n
¡a
n
=
b ¡a=over2
n
.Let d =
bn¡an
2
.If f(d) = 0,then we have found a < d < b with
f(d) = 0 and are done.If f(d) > 0,let a
n+1
= a
n
,b
n+1
= d.If f(d) < 0,let
a
n+1
= d and
Let I
n+1
= [b
n+1
;a
n+1
].Then I
n+1
½ I
n
,f(a
n+1
) < 0 < f(b
n+1
) and
b
n+1
¡a
n+1
=
b¡a
2
n
.
By the nested interval theorem,there is c 2
T
1n=0
I
n
.We claim that
f(c) = 0.
Since a
n
;c 2 I
n
,ja
n
¡ cj ·
b¡a
2
n
for all n 2 N.If ² > 0,choose N such
that
b¡a
2
N
< ².Then ja
n
¡cj < ² for all n ¸ N.Hence (a
n
)
1n=1
converges to c.
Thus,by Lemma 2,(f(a
n
))
1n=1
converges to f(c).Since f(a
n
) · 0 for all n,
we must have f(c) · 0.
Similarly,(b
n
)
1n=1
!c and (f(b
n
))
1n=1
!f(c).But each f(b
n
) > 0,thus
f(c) ¸ 0.Hence f(c) = 0.
Thus there is a < c < b with f(c) = 0.
4