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Every continuous function $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous on every bounded set.

Here's what I did so far. Let $U \subset \mathbb{R}$ be bounded set. Let $\epsilon> 0$. For every $x \in U$, there is a $\delta_x$ such that if $|a - x| < \delta_x$, then $|f(x) - f(a)| < \epsilon$. Now here's, where I'm not sure if my reasoning is correct. We have

$$U \subseteq \bigcup_{x \in U} (x - \delta_x, x + \delta_x) $$

It seems intuitively clear that since $U$ is bounded, the union of only finitely many such intervals contains $U$, in which case we can simply let $\delta = \min\{\delta_x : x \in I\}$, where $I$ is a finite indexing set.

Help please?

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2  
$U$ isn't necessarily compact. –  12F8916 May 31 '13 at 22:13
4  
... but the closure of $U$ is. –  Hagen von Eitzen May 31 '13 at 22:14
    
@HagenvonEitzen So do I just do the same argument for $cl(U)$, (proving uniform continuity on closure) and say since $U \subseteq cl(U)$, the result follows? –  user80464 May 31 '13 at 22:16
1  
@user80464 Yes. Note that for a non-compact $U$, finitely many of you intervals will not suffice in general, for example $$ (0,1) \subseteq \bigcup_{x\in (0,1)} \left(\frac x2, \frac {3x}2\right) $$ –  martini May 31 '13 at 22:21
    
@DonAntonio Yes, that's why the title says $f:{\color{red}{\Bbb R}}\to\Bbb R$. –  Pedro Tamaroff Jun 1 '13 at 0:34

1 Answer 1

There is an infinite number of items in your sum, so you can't take the min so easily. You could have $\delta=0$. You have the good idea, but we need to construct the finite sum. We will assume U closed, because we can always do the same on the closure of U.

Given $\epsilon>0$ and $x_0=\min U$, we will define $\forall n,\ x_n=\max\left\{ { x\in \left[x_{n-1}, \infty \right] } \mid { \lvert f(x)-f(x_{n-1})\rvert\leq\epsilon } \right\}$. Note that $x_n$ can be equal to $\infty$. $\left(x_n\right)$ is stritly incresing, so converge to $x_\infty \in \mathbb R_+ \cup \{\infty\}$. If $x_\infty \neq \infty$, then $f$ is not continuous in $x_\infty$, which is not possible.

We have $U $ bounded, and $x_n \rightarrow +\infty$, so $\exists n_0, \forall n>n_0, x_{n} \notin U $. Now $U \subset \bigcup_{i=0}^{n_0} [x_i, x_{i+1}]$ and you have your finite sum.

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I think you mean "union", not sum. –  40 votes Jul 14 '13 at 1:10

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