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Math Help - 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 f if is uniformly continuous on the inteval (a,b) then it is bounded on the interval.

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

    Basically if f isn't bounded, then there must be a point c in (a,b)such that at c 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 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 \left\{x_n\right\} such that |f(x_n)|\geq n.
    You can extract from \left\{x_n\right\} a Cauchy subsequence since it's a bounded one. |f| is also uniformly continuous hence \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 (a,b)=(0,1). We apply the definition of uniform continuity for \varepsilon =1. We get a \delta such that if x,y\in (0,1) and |x-y|\leq \delta then |f(x)-f(y)|\leq 1. We choose an integer n_0 such that \dfrac 1{n_0}\leq \delta. We put x_j := \dfrac j{n_0} for j\in\left\{1,\ldots,n_0-1\right\}. For x\in (0,1), exists j \in\left\{1,\ldots,n_0-1\right\} such that |x-x_j|\leq\dfrac 1{n_0}\leq \delta. Hence we get \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 x.
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  5. #5
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    Quote Originally Posted by girdav View Post
    For x\in (0,1), exists j \in\left\{1,\ldots,n_0-1\right\} such that |x-x_j|\leq\dfrac 1{n_0}\leq \delta. Hence we get \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 x.
    I'm not sure I understand why |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 |x-x_j|\leq\dfrac 1{n_0}\leq \delta necessarily...
    Given \delta  > 0, we can find a positive integer such that \frac{1}{n_0}<\delta.
    Look at the collection x_j=\frac{j}{n_0},~j=0,1,\cdots,n.
    They partition the interval [0,1] into subintervals of length \frac{1}{n_0}.
    So if x\in (0,1) it is in one of the subintervals.
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