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Thread: closed interval continiuous

  1. #1
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    closed interval continiuous

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

    Assume for contradiction that $\displaystyle f $ is not uniformly continuous. Then there are two equivalent sequences $\displaystyle (x_{n})_{n=0}^{\infty} $ and $\displaystyle (y_{n})_{n=0}^{\infty} $ in $\displaystyle [a,b] $ such that the sequences $\displaystyle (f(x_{n}))_{n=0}^{\infty} $ and $\displaystyle (f(y_{n}))_{n=0}^{\infty} $ are not equivalent. Then we let $\displaystyle 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?
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  2. #2
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    If $\displaystyle f$ is not uniformly continous then for there is $\displaystyle \epsilon > 0$ such that for any $\displaystyle \delta > 0$ we have $\displaystyle |x-y|<\delta$, $\displaystyle x,y\in [a,b]$ but $\displaystyle |f(x)-f(y)|\geq \epsilon$.
    Let $\displaystyle \delta = \frac{1}{n}$ for $\displaystyle n\in \mathbb{Z}^+$.
    Then we form two sequences $\displaystyle \{x_n\}$ and $\displaystyle \{y_n\}$ such that $\displaystyle |x_n-y_n|<\tfrac{1}{n}$ and $\displaystyle |f(x_n)-f(y_n)| \geq \epsilon$. Since $\displaystyle \{x_n\}$ is a bounded sequence by Bolzano-Weierstrass theorem there is a convergent subsequence $\displaystyle X_n$. Let $\displaystyle X = \lim ~ X_n$ and note that $\displaystyle X\in [a,b]$. Let $\displaystyle Y_n$ be the subsequence chosen in the same manner as $\displaystyle X_n$ i.e. if $\displaystyle X_n = x_{g(n)}$ for some choice function $\displaystyle g$ then $\displaystyle Y_n = y_{g(n)}$. Now since $\displaystyle |X_n - Y_n|<\tfrac{1}{g(n)}\leq \tfrac{1}{n}$ we see that $\displaystyle X_n - \tfrac{1}{n} < Y_n < X_n + \tfrac{1}{n}$. But each $\displaystyle X_n$ converges to $\displaystyle X$ and so by squeeze-theorem $\displaystyle Y_n$ converges to $\displaystyle X$. Thus, $\displaystyle \{X_n\}$ and $\displaystyle \{Y_n\}$ are two convergent sequences to $\displaystyle X\in [a,b]$. By continuity it means $\displaystyle \{f(X_n)\}$ and $\displaystyle \{f(Y_n)\}$ converge both to $\displaystyle f(X)$. However, $\displaystyle |f(X_n)-f(Y_n)| \geq \epsilon$ and this is a contradiction.
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  3. #3
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    And the set $\displaystyle E $ is infinite? Because if it was finite, then there would be two sequences that are equivalent?
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