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ureji1c8r1

ureji1c8r1

Answered question

2022-04-07

Suppose that f : [ a , b ] R is continuous and that f ( a ) < 0 and f ( b ) > 0. By the intermediate-value theorem, the set S = { x [ a , b ] : f ( x ) = 0 } is nonempty. If c = sup S, prove that c S.

My first thought was to show that S is finite therefore c S, but there is no guarantee that f doesn't have infinitely many zeros.

A thought I have now is to show that c > max ( S ) can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.

Answer & Explanation

Nollothwnfcm

Nollothwnfcm

Beginner2022-04-08Added 12 answers

You've noted that when S is finite, then c = sup S S .

Now what if S is infinite? Then there exists a sequence in S that converges to sup S .. (There are several ways to prove this. Try it on your own if you don't already know how. But a standard proof would be to choose c 0 , then choose c 1 such that it's bigger than c 0 but less than c. We know such a number must exist because if not, c 0 would be a smaller upper bound than c. And we continue on in this fashion.) Call this sequence c n . Now as f is continuous,
f ( c ) = lim n f ( c n ) .
Thus f ( c ) = 0.

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