1. Chapter 5
Limits and Continuity

2. Properties of Continuous Functions
Section 5.3
Properties of Continuous Functions

3. The continuous image of a closed set may not be closed.
The continuous image of a bounded set may not be bounded.
Note: A function f : D  is said to be bounded if its range f (D) is
a bounded subset of
So, a continuous function may not be bounded, even if its domain is bounded.
But if a set is both closed and bounded, then its image under a continuous
function will be both closed and bounded.
The Heine-Borel Theorem says S is compact iff S is closed and bounded.
So, the continuous image of a compact set is compact.
Recall from Definition 3.5.1
A set S is compact if every open cover of S contains a finite subcover.

4. Theorem 5.3.2
Let D be a compact subset of and suppose f : D  is continuous.
Then f (D) is compact.
Idea of proof:
Let G = {G} be an open cover of f (D).
Since f is continuous, the pre-image of each set in G is open.
These pre-images form an open cover of D.
Since D is compact, there is a finite subcollection that covers D.
The corresponding sets in the range form a finite subcovering of f (D).
Hence f (D) is compact.
Open cover of D
Open cover of f (D) f D
f (D)

5. Corollary 5.3.3
Let D be a compact subset of and suppose f : D  is continuous.
Then f assumes minimum and maximum values on D. That is, there exist
points x1 and x2 in D such that f (x1)  f (x)  f (x2) for all x  D.
Proof:
We know from Theorem that f (D) is compact.
Lemma tells us that f (D) has both a minimum, say y1, and a maximum, say y2.
Since y1 and y2 are in f (D), there exist x1 and x2 in D such that f (x1) = y1 and f (x2) = y2.
It follows that f (x1)  f (x)  f (x2) for all x  D. 
If we require the domain of the function to be a compact interval, then the
following holds:

6. Lemma 5.3.5
Let f : [a, b]  be continuous and suppose f (a) < 0 < f (b). Then there
exists a point c in (a, b) such that f (c) = 0.
Proof:
Let S = {x  [a, b] : f (x)  0}.
Since a  S, S is nonempty.
Clearly, S is bounded above by b, and we may let c = sup S.
We claim f (c) = 0.
Suppose f (c) < 0.
Then there exists a neighborhood U of c such that f (x) < 0 for all
x  U  [a, b].
And U contains a point p such that c < p < b.
y = f (x)
But f ( p) < 0, since p  U, so p  S.
This contradicts c being an upper bound for S. S U   a b c p

7. Lemma 5.3.5
Let f : [a, b]  be continuous and suppose f (a) < 0 < f (b). Then there
exists a point c in (a, b) such that f (c) = 0.
Proof:
Let S = {x  [a, b] : f (x)  0}.
Since a  S, S is nonempty.
Clearly, S is bounded above by b, and we may let c = sup S.
We claim f (c) = 0.
Now suppose f (c) > 0.
Then there exists a neighborhood U of c such that
f (x) > 0 for all x  U  [a, b].
And U contains a point p such that a < p < c.
y = f (x)
Since f ( x) > 0 for all x in U, S  [p, c] = .
This implies that p is an upper bound for S, and
contradicts c being the least upper bound for S. S U   a b p c
We conclude that f (c) = 0. 
The Intermediate Value Theorem extends this to a more general setting.

8. Theorem 5.3.6 (Intermediate Value Theorem)
Suppose f : [a, b]  is continuous. Then f has the intermediate value property
on [a, b]. That is, if k is any value between f (a) and f (b) , i.e.
f (a) < k < f (b) or f (b) < k < f (a),
then there exists a point c in (a, b) such that f (c) = k.
Theorem 5.3.6
Proof:
Suppose f (a) < k < f (b). Apply Lemma to the continuous function
g : [a, b]  given by g(x) = f (x) – k.
Then g(a) = f (a) – k < 0 and g(b) = f (b) – k > 0.
Thus there exists c  (a, b) such that g(c) = 0.
But then f (c) – k = 0 and f (c) = k.
When f (b) < k < f (a), a similar argument applies to the function g(x) = k – f (x). 

9. Theorem 5.3.6 (Intermediate Value Theorem)
Suppose f : [a, b]  is continuous. Then f has the intermediate value property
on [a, b]. That is, if k is any value between f (a) and f (b) , i.e.
f (a) < k < f (b) or f (b) < k < f (a),
then there exists a point c in (a, b) such that f (c) = k.
Theorem 5.3.6
The idea behind the intermediate value theorem is simple when we graph the function.
y = f (x)
If k is any height between f (a) and f (b), a b
f (b)
there must be some point c in (a, b) such
that f (c) = k.
y = k
That is, if the graph of f is below y = k
at a and above y = k at b, then for f to be
continuous on [a, b], it must cross y = k
somewhere in between.
f (a)
The graph of a continuous function can
have no “jumps.” c

10. C Example 5.3.8 (an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Given any   [0, 2], let r be a ray from the origin having angle  with the positive x axis.
This ray determines a unique circumscribing rectangle whose sides are parallel and
perpendicular to r.
Let A( ) be the length of the sides parallel to r.
A( )
Let B( ) be the length of the sides
perpendicular to r.
B( ) C
Now define f : [0, 2]  by
f ( ) = A( ) – B( ).
If for some  the circumscribing rectangle
is not a square, then A( )  B( ), and we
suppose f ( ) = A( ) – B( ) > 0. r 

11. C Example 5.3.8 (an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Given any   [0, 2], let r be a ray from the origin having angle  with the positive x axis.
We have f ( ) = A( ) – B( ) > 0.
Now let  =  +  /2.
Then the circumscribing rectangle
is unchanged except the labeling
of its sides is reversed.
B( )
A( )
B( )
So f ( ) = A( ) – B( ) < 0. C
A( )
Assuming that f is a continuous function,
the Intermediate Value Theorem implies
there exists some angle  with  <  < 
such that f ( ) = 0. r 
But then A( ) = B( ) and the circumscribing
rectangle for this angle is a square. 

12. C Example 5.3.8 (an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Example 5.3.8 C  

13. C Example 5.3.8 (an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Example 5.3.8 C  

14. C Example 5.3.8 (an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Example 5.3.8 C  

15. C Example 5.3.8 Now the circumscribing rectangle
(an application in geometry)
Let C be a bounded closed subset of the plane. We claim there exists a square S that
circumscribes C. That is, C  S and each side of S touches C.
Now the circumscribing rectangle
has become a square, with  <  <  . C   