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# boundedness theorem

Boundedness Theorem. Let $a$ and $b$ be real numbers with $a<b$, and let $f$ be a continuous, real valued function on $[a,b]$. Then $f$ is bounded above and below on $[a,b]$.

Proof. Suppose not. Then for all natural numbers $n$ we can find some $x_{n}\in[a,b]$ such that $|f(x_{n})|>n$. The sequence $(x_{n})$ is bounded, so by the Bolzano-Weierstrass theorem it has a convergent sub sequence, say $(x_{{n_{i}}})$. As $[a,b]$ is closed $(x_{{n_{i}}})$ converges to a value in $[a,b]$. By the continuity of $f$ we should have that $f(x_{{n_{i}}})$ converges, but by construction it diverges. This contradiction finishes the proof.

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## Mathematics Subject Classification

26A06*no label found*

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new question: Prime numbers out of sequence by Rubens373

Oct 7

new question: Lorenz system by David Bankom

Oct 19

new correction: examples and OEIS sequences by fizzie

Oct 13

new correction: Define Galois correspondence by porton

Oct 7

new correction: Closure properties on languages: DCFL not closed under reversal by babou

new correction: DCFLs are not closed under reversal by petey

Oct 2

new correction: Many corrections by Smarandache

Sep 28

new question: how to contest an entry? by zorba

new question: simple question by parag

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(v6) by classicleft 2013-03-22