closed interval continiuous

**particlejohn** · August 6th 2008, 08:35 AM

Prove the following: Let $a < b$ be real numbers. and let $f: [a,b] \to \bold{R}$ be a function which is continuous on $[a,b]$ . Then $f$ is also uniformly continuous.

Assume for contradiction that $f$ is not uniformly continuous. Then there are two equivalent sequences $(x_{n})_{n=0}^{\infty}$ and $(y_{n})_{n=0}^{\infty}$ in $[a,b]$ such that the sequences $(f(x_{n}))_{n=0}^{\infty}$ and $(f(y_{n}))_{n=0}^{\infty}$ are not equivalent. Then we let $E = \{n \in \bold{N}: f(x_{n}) \ \text{and} \ f(y_{n}) \ \text{are not} \ \varepsilon \ \text{close} \}$ . Then use Bolzano-Weierstrass Theorem to obtain contradiction?

Is this correct?

**ThePerfectHacker** · August 6th 2008, 11:38 AM

If $f$ is not uniformly continous then for there is $\epsilon > 0$ such that for any $\delta > 0$ we have $|x-y|<\delta$ , $x,y\in [a,b]$ but $|f(x)-f(y)|\geq \epsilon$ .
Let $\delta = \frac{1}{n}$ for $n\in \mathbb{Z}^+$ .
Then we form two sequences $\{x_n\}$ and $\{y_n\}$ such that $|x_n-y_n|<\tfrac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \epsilon$ . Since $\{x_n\}$ is a bounded sequence by Bolzano-Weierstrass theorem there is a convergent subsequence $X_n$ . Let $X = \lim ~ X_n$ and note that $X\in [a,b]$ . Let $Y_n$ be the subsequence chosen in the same manner as $X_n$ i.e. if $X_n = x_{g(n)}$ for some choice function $g$ then $Y_n = y_{g(n)}$ . Now since $|X_n - Y_n|<\tfrac{1}{g(n)}\leq \tfrac{1}{n}$ we see that $X_n - \tfrac{1}{n} < Y_n < X_n + \tfrac{1}{n}$ . But each $X_n$ converges to $X$ and so by squeeze-theorem $Y_n$ converges to $X$ . Thus, $\{X_n\}$ and $\{Y_n\}$ are two convergent sequences to $X\in [a,b]$ . By continuity it means $\{f(X_n)\}$ and $\{f(Y_n)\}$ converge both to $f(X)$ . However, $|f(X_n)-f(Y_n)| \geq \epsilon$ and this is a contradiction.

**particlejohn** · August 6th 2008, 11:41 AM

And the set $E$ is infinite? Because if it was finite, then there would be two sequences that are equivalent?

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