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- The set is a bounded set. Hence, its least upper bound exists by least upper bound property of the real numbers. Let on . If there is no point x on [a, b] so that f = M, then
on [a, b]. Therefore, is continuous on [a, b].
However, to every positive number ε, there is always some x in [a, b] such that because M is the least upper bound. Hence, , which means that is not bounded. Since every continuous function on [a, b] is bounded, this contradicts the conclusion that was continuous on [a, b]. Therefore, there must be a point x in [a, b] such that f = M.
∎ (en)
- In the setting of non-standard calculus, let N  be an infinite hyperinteger. The interval [0, 1] has a natural hyperreal extension. Consider its partition into N subintervals of equal infinitesimal length 1/N, with partition points xi = i /N as i "runs" from 0 to N. The function ƒ  is also naturally extended to a function ƒ* defined on the hyperreals between 0 and 1. Note that in the standard setting , a point with the maximal value of ƒ can always be chosen among the N+1 points xi, by induction. Hence, by the transfer principle, there is a hyperinteger i0 such that 0 ≤ i0 ≤ N and   for all i = 0, ..., N. Consider the real point
where st is the standard part function. An arbitrary real point x lies in a suitable sub-interval of the partition, namely , so that  st = x. Applying st to the inequality , we obtain . By continuity of ƒ  we have
:.
Hence ƒ ≥ ƒ, for all real x, proving c to be a maximum of ƒ.
∎ (en)
- By the Boundedness Theorem, is bounded above on and by the completeness property of the real numbers has a supremum in . Let us call it , or . It is clear that the restriction of to the subinterval where has a supremum which is less than or equal to , and that increases from to as increases from to .
If then we are done. Suppose therefore that and let . Consider the set of points in such that .
Clearly ; moreover if is another point in then all points between and also belong to because is monotonic increasing. Hence is a non-empty interval, closed at its left end by .
Now is continuous on the right at , hence there exists such that for all in . Thus is less than on the interval so that all these points belong to .
Next, is bounded above by and has therefore a supremum in : let us call it . We see from the above that . We will show that is the point we are seeking i.e. the point where attains its supremum, or in other words .
Suppose the contrary viz. . Let and consider the following two cases:
# . As is continuous at , there exists such that for all in . This means that is less than on the interval . But it follows from the supremacy of that there exists a point, say, belonging to which is greater than . By the definition of , . Let then for all in , . Taking to be the minimum of and , we have for all in . Hence so that . This however contradicts the supremacy of and completes the proof.
# . As is continuous on the left at , there exists such that for all in . This means that is less than on the interval . But it follows from the supremacy of that there exists a point, say, belonging to which is greater than . By the definition of , . Let then for all in , . Taking to be the minimum of and , we have for all in . This contradicts the supremacy of and completes the proof. ∎ (en)
- By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound M of f exists. It is necessary to find a point d in [a, b] such that M = f. Let n be a natural number. As M is the least upper bound, M – 1/n is not an upper bound for f. Therefore, there exists dn in [a, b] so that M – 1/n n}. Since M is an upper bound for f, we have M – 1/n < f ≤ M for all n. Therefore, the sequence {f} converges to M.
The Bolzano–Weierstrass theorem tells us that there exists a subsequence {}, which converges to some d and, as [a, b] is closed, d is in [a, b]. Since f is continuous at d, the sequence {f} converges to f. But {f} is a subsequence of {f} that converges to M, so M = f. Therefore, f attains its supremum M at d''.
∎ (en)
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