# Class notes Nov ```Class notes Nov. 13
The Intermediate Value Theorem
We start by stating the Intermediate Value Theorem (Theorem 4.5.1. in Abbott) and give
its proof.
Theorem 1. Let f :[a, b]  R be a continuous function and let L be a real number
satisfying f (a)  L  f (b) or f (a)  L  f (b) . Then there exists a point a  c  b such
that f (c)  L .
Proof: Let us assume for convenience that f (a)  L  f (b) . Define the function
g ( x)  f ( x)  L .
Then g is continuous on [a, b] , and g (a)  f (a)  L  0, g (b)  f (b)  L  0 . Then the
conclusion of the theorem is equivalent to showing that g must be zero somewhere inside
the interval [a, b] .
Define the following sets: A1  {x [a, b]: g ( x)  0}, A2  {x [a, b]: g ( x)  0} . Both sets
are bounded, so by the Axiom of Completeness they both have a supremum and an
infimum. Let a1  inf A1 , b1  sup A1 , a2  inf A2 , b2  sup A2 . Notice that a  A1 , b  A2 , so
inf A1  a,sup A2  b .
Since b1  sup A1 there exists a sequence ( xn )  A1 : lim xn  b1 . But since the function g is
n 
continuous, we have lim g ( xn )  g (b1 ) , and since g ( xn )  0, for all n  1, g (b1 )  0 ,
n 
therefore b1  b . This allows us to construct a sequence converging to b1 from above:
yn  b1 : lim yn  b1 . Since b1  sup A1 we have
g ( yn )  0, for all n  1 and
n 
lim g ( yn )  g (b1 )  0 . So we found a point a  b1  sup A1  b such that both g (b1 )  0
n 
and g (b1 )  0 must be true. Then g (b1 )  0 and the proof is complete.
QED
Following you will find some hints to some of the homework problems due Nov15.
Exercise 4.4.4. We are given f :[a, b]  R, f ( x)  0 on [a, b] . Now, for the function
to be bounded on [a, b] , we need to show that there exists M  0 :
have that
f ( x)  0
so
1
 0 . The statement
f ( x)
1
f
1
f ( x)
1
M
f ( x)
is equivalent to
1
 0 , so what you need to show is that the function has a positive lower bound
M
on the interval [a, b] . Can you say anything about inf{ f ( x) : x  [a, b]} ?
For Exercise 4.4.8 use the approach in Exercise 4.4.7 below.
f ( x) 
Exercise 4.4.7. We have g uniformly continuous on (a, b] and [b,c),a&lt;b&lt;c .
We write the uniform continuity for the two intervals. Let   0 . We know that there

exists 1  0 such that for any x, y  (a, b] :| x  y | 1 ,| g ( x)  g ( y ) | (1). You will see
2

later why we chose
in the above.
2
Also, since g is uniformly continuous on [b,c) , there exists  2  0 such that for any

x, y  [b, c) :| x  y |  2 ,| g ( x)  g ( y ) | (2).
2
Let   min{1 ,  2 } . From (1) and (2) it follows that if x, y  (a, c) :| x  y |  and either

x, y  (a, b] or x, y  [b, c) then | g ( x)  g ( y ) | . However, we should ask ourselves
2
what happens if x  b  y :| x  y |  ? Notice that in this case | x  b |  and | y  b | 


and | g ( x)  g (b) | ,| g (b)  g ( y ) | . Then we write
2
2
 
| g ( x)  g ( y ) || g ( x)  g (b) |  | g (b)  g ( y ) |   
2 2
and we are done.
Remark. If in the above we knew that the function is uniformly continuous on two
overlapping intervals, such as (a, b],[d , c), a  d  b  c then there would be no need of
discussing the case x  b  y :| x  y |  separately. Think about this when doing
exercise 4.4.8.
Exercise 4.4.11. (Topological Characterization of Continuity) Let g : R  R . Define the
preimage of a set A: g 1 ( A)  {x  R : g ( x)  A} . Show that g is continuous if and only if
for any open set O  R, g 1 (O) is open.
Proof: We need to prove two implications here.
Let’s assume first that g is continuous, and let O  R be open. We will prove that
g 1 (O) is open by proving that all points in g 1 (O) are interior points. Let a  g 1 (O) .
Then g (a)  O and since O is open, there exists a neighborhood of g ( a ) contained in O,
that is: there exists   0 such that ( g (a)   , g (a)   )  O . Since the function is
continuous at a there exists   0 such that for any x :| x  a |  ,| g ( x)  g (a) |  , or
equivalently g ( x)  ( g (a)   , g (a)   )  O . But if g ( x)  O it follows that x  g 1 (O)
for all x :| x  a |  , so (a   , a   )  g 1 (O) , proving that g 1 (O) is open.
Conversely, let us assume g 1 (O) is open for any open set O. Let a  R and let   0 .
Define the open set O  ( g (a)   , g (a)   ) . According to our assumption g 1 (O) must
be open as well, and we also know that a  g 1 (O) . Since g 1 (O) is open we can find a
 -neighborhood of a contained in g 1 (O) , that is there exists   0 such that
(a   , a   )  g 1 (O) . This last statement implies that g ( x)  O for all x :| x  a |  , or
equivalently | g ( x)  g (a) |  for all x :| x  a |  , proving that g is continuous at a.
QED
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