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February 21st, 2013, 05:44 PM   #1
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intermediate value theorem

For the following proof of the intermediate value theorem ,which i found in wikipedia:

Proof:
Let S be the set of all x in [a, b] such that f(x) ? u. Then S is non-empty since a is an element of S, and S is bounded above by b. Hence, by completeness, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f(c) = u.
Suppose first that f(c) > u, then f(c) ? u > 0. Since f is continuous, there is a ? > 0 such that |?f(x) ? f(c)?| < ? whenever |?x ? c?| < ?. Pick ? = f(c) ? u, then |?f(x) ? f(c)?| < f(c) ? u. But then, f(x) > f(c) ? (f(c) ? u) = u whenever |?x ? c?| < ? (that is, f(x) > u for x in (c ? ?, c + ?)). This requires that c ? ? be an upper bound for S (since no point in the interval (c ? ?, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption.
Suppose instead that f(c) < u. Again, by continuity, there is a ? > 0 such that |?f(x) ? f(c)?| < u ? f(c) whenever |?x ? c?| < ?. Then f(x) < f(c) + (u ? f(c)) = u for x in (c ? ?, c + ?). Since x=c + ?/2 is contained in (c ? ?, c + ?), it also satisfies f(x) < u, so it must be contained in S. However, it also exceeds the least upper bound c of S. The contradiction nullifies this paragraph's opening assumption, as well.

We deduce that f(c) = u as stated.

I have the folowing questions:

1)How does the author come to the conclusion that ,c-? is an upper bound for S ,in this part of the proof:

"This requires that c ? ? be an upper bound for S (since no point in the interval (c ? ?, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption"

2) How does the author conclude that ,x = c+?/2 is contained in S ,since he has not proved that in the following part of the proof:
"Suppose instead that f(c) < u. Again, by continuity, there is a ? > 0 such that |?f(x) ? f(c)?| < u ? f(c) whenever |?x ? c?| < ?. Then f(x) < f(c) + (u ? f(c)) = u for x in (c ? ?, c + ?). Since x=c + ?/2 is contained in (c ? ?, c + ?), it also satisfies f(x) < u, so it must be contained in S"

3) Nowhere in the proof proves that : a<c<b
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