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The function H is continuous on the compact space K and therefore has a mininimum. Since
H y 0 for every y K there must be a positive number such that H y
for every y H. 228 For such a number we must have d x, y for all x H and all y K. 12. a. Suppose that S is a compact subspace of a metric space X and that f is a continuous function from X to
R. Suppose that c is a point of the set S, that f c 1 and that f x 1 for all x X S. Prove that the
function f has a minimum.
On the compact space S, the function f must have a minimum. Since f c 1 we know that the
minimum value of f on S cannot exceed 1. Since f x 1 whenever x X S, we conclude that
the minimum value of f on the subspace S is the minimum value of f on the whole space X.
b. Prove that if f is a quartic polynomial of the form
f x x 4 ax 3 bx 2 cx d
for every number x then the function f must have a minimum.
Since
f x x 4 1 a b c3 d
x
x2
x
x4
for all x 0, it is clear that f x
Ý as n Ý. Using this fact we choose a number p 0 such
that f x d whenever x  p.
p
0
p
The minimum value that f has on the interval p, p is the minimum of f on the entire line R. Exercises on Continuity of Functions on Intervals
1. Given that S is a set of positive numbers and that f x x for all x S, prove that f is a oneone continuous
function on S. Prove that S is an interval if and only if the set f S is an interval. Hint: Once you have shown that f is strictly increasing and continuous on S, the fact that S is an
interval if and only if f S is an interval will follow from the Bolzano intermediate value theorem and
this theorem.
2. Prove that there are three real numbers x satisfying the equation
x 3 4x 2 0. Solution: First look at a sketch of the graph y x 3 4x 2 4 2
2 1x 1 2 0
2 4 We now define f x x 3 4x 2 for every real number x and observe that f 2 0 and f 1 0.
Therefore Bolzano’s intermediate theorem guarantees that the equation x 3 4x 2 0 has a solution
between 2 and 1. Since f 1 0 and f 0 0 we know that there is a solution of the equation between
1 and 0. Finally, from the fact that f 0 0 and f 3 0 we know that there is a solution of the equation
between 0 and 3.
3. Is it true that if a set S of real numbers is not an interval then there must exist a oneone continuous function
on S whose inverse function fails to be continuous?
Not at all. In fact, we know that if S is closed and bounded then every oneone continuous function
S must have a continuous inverse function. Look also at the case in which S is the union of two
open intervals that do not intersect with each other. If f is a oneone continuous function on S then 229 the range of f is also the union of two open intervals that do not intersect with each other and the
inverse function of f will be continuous.
4. Is it true that if a set S of real numbers is not an interval and is not closed then there must exist a oneone
continuous function on S whose inverse function fail...
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 Fall '08
 STAFF
 Math, Calculus

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