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Thread: Proof that a uniformly continuous function on an open interval is bounded

  1. #1
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    Proof that a uniformly continuous function on an open interval is bounded

    I'm trying to prove that if $\displaystyle f$ if is uniformly continuous on the inteval $\displaystyle (a,b)$ then it is bounded on the interval.

    I thought to prove it by contradiction - assuming that there exists a uniformly continuous function $\displaystyle f$ on $\displaystyle (a,b)$ that is not bounded, and see that that leads to a contradiction.

    Basically if $\displaystyle f$ isn't bounded, then there must be a point $\displaystyle c$ in $\displaystyle (a,b)$such that at $\displaystyle c$ $\displaystyle lim f$ = infinity. Therefore in any the neighborhood of the limit we can always find two values of the function which are very far away from each other, contradicting the assumption that $\displaystyle f$ is uniformly continuous.

    Is this the right direction? And if it is, how do I got about formalizing it?

    Thanks in advance.
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  2. #2
    Super Member girdav's Avatar
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    It's the idea indeed. A point on which the limit is infinite can only be a or b.
    Assume that f is not bounded. We can find a sequence $\displaystyle \left\{x_n\right\}$ such that $\displaystyle |f(x_n)|\geq n$.
    You can extract from $\displaystyle \left\{x_n\right\}$ a Cauchy subsequence since it's a bounded one. $\displaystyle |f|$ is also uniformly continuous hence $\displaystyle \left\{|f(x_n)|\right\}$ is also a Cauchy sequence. We get a contradiction.
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    I'm afraid I haven't learned anything about sequences yet...is there another way?
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    Super Member girdav's Avatar
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    We will show the result in the case $\displaystyle (a,b)=(0,1)$. We apply the definition of uniform continuity for $\displaystyle \varepsilon =1$. We get a $\displaystyle \delta$ such that if $\displaystyle x,y\in (0,1)$ and $\displaystyle |x-y|\leq \delta$ then $\displaystyle |f(x)-f(y)|\leq 1$. We choose an integer $\displaystyle n_0$ such that $\displaystyle \dfrac 1{n_0}\leq \delta$. We put $\displaystyle x_j := \dfrac j{n_0}$ for $\displaystyle j\in\left\{1,\ldots,n_0-1\right\}$. For $\displaystyle x\in (0,1)$, exists $\displaystyle j \in\left\{1,\ldots,n_0-1\right\}$ such that $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$. Hence we get $\displaystyle \displaystyle|f(x)| \leq |f(x)-f(x_j)|+|f(x_j)|\leq 1+\max_{1\leq j\leq n_0-1}|f(x_j)|$, a bound which doesn't depend on $\displaystyle x$.
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  5. #5
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    Quote Originally Posted by girdav View Post
    For $\displaystyle x\in (0,1)$, exists $\displaystyle j \in\left\{1,\ldots,n_0-1\right\}$ such that $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$. Hence we get $\displaystyle \displaystyle|f(x)| \leq |f(x)-f(x_j)|+|f(x_j)|\leq 1+\max_{1\leq j\leq n_0-1}|f(x_j)|$, a bound which doesn't depend on $\displaystyle x$.
    I'm not sure I understand why $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$ necessarily...
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  6. #6
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    Quote Originally Posted by moses View Post
    I'm not sure I understand why $\displaystyle |x-x_j|\leq\dfrac 1{n_0}\leq \delta$ necessarily...
    Given $\displaystyle \delta > 0$, we can find a positive integer such that $\displaystyle \frac{1}{n_0}<\delta$.
    Look at the collection $\displaystyle x_j=\frac{j}{n_0},~j=0,1,\cdots,n$.
    They partition the interval $\displaystyle [0,1]$ into subintervals of length $\displaystyle \frac{1}{n_0}$.
    So if $\displaystyle x\in (0,1)$ it is in one of the subintervals.
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