Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Firstly, I haven't done any math in 4 months so I am bit rusty.

Is there a difference between the boundedness of a continuous function depending on whether we are taking it over a closed or open interval?

My view: If the interval is open, say $I = (0, 1)$ then a continuous function can be unbounded. $1/x$ for example has no upper bound near 0. So in this case it seems straightforward enough?

If the interval is closed, say $I = [0, 1]$ then it seems to me that the same situation applies, there is no upper bound as a cts function can go as 'high' as it wants...However it has to be defined at 0 and 1 for it to be continuous so it doesn't go to infinity...it will always attain a real value at 0 or 1. This seems to mean that it does have to have an upper bound.

Can anybody clear this up for me?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

A continuous function on a closed interval is bounded and attains its bounds, by the Extreme Value Theorem. The example you give for open intervals is correct, so the "closed" assumption is necessary.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.