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?
Best Answer
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.