Continuous function on intervals

[rapple]

Homework Statement

f:(0,1]->R be a continuous function. Is it possible that f does not have an absolute min or max. Give counter examples

Homework Equations

The Attempt at a Solution

Since f is partially bounded, if I break the interval down into smaller sub intervals, each will have an abs min and max. So I will have at least one of the extremes.

 

[Dick]

The problem says 'give counterexamples'. Do that. They aren't hard to find.

 

[VeeEight]

"if I break the interval down into smaller sub intervals, each will have an abs min and max."

If you break the interval into a union smaller intervals, you will always have one interval of the form (0, a) where a is some real in (0,1]. What is the difference between [0,1] and (0,1]? Does a continuous function on [0,1] attain a min/max?

Think of some functions on (0,1] that may not have a min or max (you might want to think of f(x) as x approaches 0)

 

[rapple]

How about f(x) = 1 if x=1 and f(x) = 1/(x-1) if x Not= 1. Then as x->0 , f(x) is increasing and as x-> 1, f(x) is decreasing. Since it is 1 at one, it is not the minimum.

 

[Dick]

That f is also not continuous on (0,1]. Keep it simple. How about f(x)=1/x?

 

[rapple]

Then f has an abs min at x=1. even though it is continuous

 

[Dick]

Then f has an abs min at x=1. even though it is continuous

Oh, I thought you wanted no max or no min. You want neither max NOR min. How about a function that oscillates with increasing increasing frequency and magnitude as x->0? Can you give an example of one of those?

 

[rapple]

How about sin(1/x)/x. But I don't think it is continuous because of the sharp peaks. Is it a form of sin function?

 

[Dick]

How about sin(1/x)/x. But I don't think it is continuous because of the sharp peaks. Is it a form of sin function?

That's perfect. It's continuous on (0,1]. The only place it has a problem is at x=0. But x=0 is not in (0,1].

 

[rapple]

Thank you. You seem to have a way of making it happen!