A real valued function is continuous on closed & bounded interval . Show that function is bounded.?

Question

Question from real analysis.

There is no ambiguity . We need to prove
mod( function) < M when function is cotinuous on interval [a , b] a,b are not infinity.

Answer 1

Most math teachers will prefer a direct proof over a proof by contradiction. This is an excellent question because both approaches give insight to the problem. When I teach analysis, I usually do both to reinforce the definition of continuity.

If f were not bounded, then for each positive integer N, there is x_N in [a,b] satisfying |f(x_N)|>N. The set {x_N} is an infinite set in [a,b] so by Bolzano-Weierstrass, It has an accumulation point c that is contained in [a,b]. Let {x_m} be a subsequence of {x_N} converging to c. Since \lim_{m \to \infty} x_m does not exist, \lim_{x \to c} f(x) does not exist, so f(x) is not continuous at c, which is a contradiction.

Your first real analysis course may be the most difficult course you take as an undergraduate. Good luck!

Challenge: Now prove that f is uniformly continuous on [a,b]. (Note that this has previously been answered in Answers and the proof will be in most analysis texts, but you should try it on your own first! :) )

Answer 2

by the heine borel theorem a closed interval can be covered by a finite set of closed intervals [x1,x2]. if you take the set of intervals [x(n-1),x(n)] to cover your interval where n-1

Answer 3

ixat02 has most of it right.
The sets defined by n-1
Then you invoke Heine-Borel to show that this covering of the interval has a finite subcovering, the function has a max.

Answer 4

I think what jxat is saying is that if the domain of a continuous function is compact on an interval, then the range of the functions value is also compact so it can't have any 'discontinuous' points (such as unboundedly large values) on that interval.


Doug

Answer 5

can you elaborate on the question please?i am afraid it is a little ambiguous