Extreme Value theorem does not hold if [a; b)

[mikael27]

Homework Statement

Show that the statement of the Extreme Value theorem does not hold if [a, b] is replaced
by [a, b).

Homework Equations

The Attempt at a Solution

Please help

 

[Dick]

It's not that hard. Please help us by at least trying.

 

[mikael27]

i just know that we are going to have open interval and that the will not be either maximum or minimum. But i don't know how to prove it

 

[Dick]

i just know that we are going to have open interval and that the will not be either maximum or minimum. But i don't know how to prove it

You don't have to prove anything. You just have to give a counterexample where the extreme value theorem doesn't hold on a half open interval. Please don't PM me. Just reply on the forum thread.

 

[mikael27]

Given that :

(Extreme value theorem). If f : [a; b] in R is continuous, then there exist c, d in [a; b] such that

f(c) = sup{f(x) | x in [a, b]};
f(d) = inf{f(x) | x in [a, b]}
Note that since c, d in [a, b], the supremum and infimum in the above two equations are in fact the maximum and minimum, respectively.

I tried the follow:

The function f : [0, 1] given by f(x) = 1 for all x in [0, 1] is continuous and for any
c,d in [0; 1] we have
f(c) = 1 = sup{f(x)| x in [0, 1]} = inf{f(x) | x in [0, 1]} = 1 = f(d):

The function f : [0, 1) in R given by f(x) = x is continuous on [0, 1). If
S = {x | x in (0, 1)};
then sup S = 1 and inf S = 0, but these values are not attained. Thus the statement does not hold if [a, b] is replaced by (a, b). 

Is this counterexample correct?

 

[Dick]

Given that :

(Extreme value theorem). If f : [a; b] in R is continuous, then there exist c, d in [a; b] such that

f(c) = sup{f(x) | x in [a, b]};
f(d) = inf{f(x) | x in [a, b]}
Note that since c, d in [a, b], the supremum and infimum in the above two equations are in fact the maximum and minimum, respectively.

I tried the follow:

The function f : [0, 1] given by f(x) = 1 for all x in [0, 1] is continuous and for any
c,d in [0; 1] we have
f(c) = 1 = sup{f(x)| x in [0, 1]} = inf{f(x) | x in [0, 1]} = 1 = f(d):

The function f : [0, 1) in R given by f(x) = x is continuous on [0, 1). If
S = {x | x in (0, 1)};
then sup S = 1 and inf S = 0, but these values are not attained. Thus the statement does not hold if [a, b] is replaced by (a, b). 

Is this counterexample correct?

Yes, but I think you could say it a lot more simply. f(x)=x on [0,1) has sup(f)=1 but there is no point x in [0,1) such that f(x)=1.

 

[mikael27]

So just say this?

The function f : [0, 1) in R given by f(x) = x is continuous on [0, 1).

then sup(f)=1 ,but there is no point x in [0,1) such that f(x)=1.

Thus the statement does not hold if [a, b] is replaced by (a, b).

 

[Dick]

So just say this?

The function f : [0, 1) in R given by f(x) = x is continuous on [0, 1).

then sup(f)=1 ,but there is no point x in [0,1) such that f(x)=1.

Thus the statement does not hold if [a, b] is replaced by (a, b).

That's good enough for me.