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.

单词	BoundednessTheorem
释义	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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