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boundedness theorem (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 $ \vert f(x_n)\vert>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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Cross-references: contradiction, diverges, converges, closed, convergent, Bolzano-Weierstrass theorem, sequence, natural numbers, proof, bounded, function, continuous, real numbers
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This is version 3 of boundedness theorem, born on 2004-07-25, modified 2004-07-25.
Object id is 6022, canonical name is BoundednessTheorem.
Accessed 4114 times total.

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AMS MSC: 26A06 (Real functions :: Functions of one variable :: One-variable calculus)
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