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- This result explains why closed bounded intervals have nicer properties than other ones.
*A continuous function on a closed bounded interval is bounded and attains its bounds.***Proof**

- Suppose
*f*is defined and continuous at every point of the interval [*a*,*b*]. Then if*f*were not bounded above, we could find a point*x*_{1}with*f*(*x*_{1}) > 1, a point*x*_{2}with*f*(*x*_{2}) > 2, ...

Now look at the sequence (*x*_{n}). By the Bolzano-Weierstrass theorem, it has a subsequence (*x*_{ij}) which converges to a point*α*∈ [*a*,*b*]. By our construction the sequence (*f*(*x*_{ij})) is unbounded, but by the continuity of*f*, this sequence*should*converge to*f*(*α*) and we have a contradiction.

The proof that*f*is bounded below is similar.To show that

*f*attains its bounds, take*M*to be the least upper bound of the set*X*= {*f*(*x*) |*x*∈ [*a*,*b*] }. We need to find a point*β*∈ [*a*,*b*] with*f*(*β*) =*M*. To do this we construct a sequence in the following way:

For each*n*∈**N**, let*x*_{n}be a point for which |*M*-*f*(*x*_{n}) | <^{1}/_{n}. Such a point must exist otherwise*M*-^{1}/_{n}would be an upper bound of*X*. Some subsequence of (*x*_{1},*x*_{2}, ... ) converges to*β*(say) and (*f*(*x*_{1}) ,*f*(*x*_{2}) , ... )→*M*and by continuity*f*(*β*) =*M*as required.

The proof that*f*attains its lower bound is similar.

**Theorem**

Here are some examples to show why you must have a

- Interval not closed

The function*f*: (0, 1]→**R**defined by*f*(*x*) =^{1}/_{x}is continuous but not bounded.

The function*f*: [0, 1)→**R**defined by*f*(*x*) =*x*is continuous and bounded but does not attain its least upper bound of 1. - Interval not bounded

The function*f*: [0, ∞)→**R**defined by*f*(*x*) =*x*is continuous but not bounded.

The function*f*: [0, ∞)→**R**defined by*f*(*x*) =^{x}/_{(1+x)}is continuous and bounded but does not attain its least upper bound of 1.

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