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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?

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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.

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