The original poster attempts to demonstrate that the Extreme Value Theorem does not hold when the interval is half-open, specifically [a, b). The discussion revolves around the implications of continuity on such intervals and the existence of maximum and minimum values.
Participants are exploring the validity of a proposed counterexample involving the function f(x) = x on the interval [0, 1). There is acknowledgment of the continuity of the function and the non-attainment of the supremum, but no explicit consensus on the final presentation of the argument has been reached.
Some participants note the need to clarify the definitions and implications of the Extreme Value Theorem, particularly regarding the nature of half-open intervals and the conditions under which the theorem applies.
mikael27 said: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
mikael27 said: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?
mikael27 said: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).