Boundedness Theorem: A continuous function on a closed interval [a,b] must be bounded on that interval.
There are two numbers - a lower bound M and an upper bound N - such that every value of f on the interval [a,b] falls between M and N.
Here are some ways a bounded function on a closed interval [a,b] could look like:
To understand why a function that's continuous on a closed interval [a,b] must be bounded on that interval, now we will think about what an unbounded function looks like. An unbounded function zooms off to infinity somewhere.
If we have a function that is unbounded on a closed interval [a,b] there are only two possibilities. The first possibility is that the function is unbounded at one of the endpoints, in which case the function is discontinuous at that endpoint:
The other possibility is that the function is unbounded somewhere in the middle of the interval, in which case the function is discontinuous somewhere in the middle of the interval:
In either case, an unbounded function on a closed interval [a,b] can't be continuous. Therefore we can't have a function on a closed interval [a,b] be both continuous and unbounded on that interval. And that means a continuous function on a closed interval [a,b] can't be unbounded (in other words, must be bounded) on that interval.
These next exercises may require a bit of thought. Try drawing some sample functions.
The Boundedness Theorem makes two assumptions and draws a conclusion. It says if we assume that both
then we can conclude that f must be bounded on that interval. The last two exercises above illustrate that we do need both assumptions. If either assumption is missing, we're not allowed to draw the conclusion, because the function won't necessarily be bounded on the interval.
In order to use the Boundedness Theorem to conclude that a function must be bounded on an interval, both of the assumptions must hold.
If it seems like the answer to the question, "Can we use the Boundedness Theorem?" is usually "no", you would be correct. We need both assumptions in order to use the theorem; if either one fails then we're out of luck.