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I've been preparing for the prelim in August, and was working on a problem involving uniform continuity and restriction of functions. I absentmindedly assumed the above by considering the contrapositive: if $f: A \rightarrow \mathbb{R}$ isn't uniformly continuous, that implies $\exists \ \epsilon$ such that no $\delta$ satisfies $d(x,y) < \delta \implies d(f(x),f(y)) < \epsilon, \,\,\ \forall x,y \in A$, and this failure of $\epsilon$'s existence shouldn't change when I "add more points" by considering $f: X \rightarrow \mathbb{R}.$

However, if this is true, we obtained a lot of results I consider to be strangely powerful. For example, if a function is continuous on $\mathbb{R}$, it is uniformly continuous on any bounded interval I, as it's uniformly continuous on $\overline{I}$ which is compact by Heine-Borel. Hence, if $f$ is a real-valued function continuous on a subset $A$ of $R$, it's uniformly continuous on any bounded subset $X$ of $A$.

Conclusions such as this seem too strong! Is there a flaw in my reasoning, and if so, where is it?

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    There is a rather difficult [theorem due to Rademacher](http://en.wikipedia.org/wiki/Rademacher's_theorem) indicating a further strong distinction between Lipschitz and uniform continuity: It asserts that a Lipschitz continuous function on $\mathbb{R}^{n}$ is differentiable "almost everywhere" (a precise technical term that should make sense intuitively), while you probably came across [examples of continuous functions on $\[0,1\]$ that aren't differentiable anywhere](http://math.stackexchange.com/q/31054/).2011-07-04

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It is true, and your conclusion that every continuous $f:\mathbb{R}\to\mathbb{R}$ is uniformly continuous on bounded subsets of $\mathbb{R}$ is also correct.

One could go further in saying precisely why it is true, which might help to convince you. Suppose $f:X\to \mathbb{R}$ is uniformly continuous and $A\subset X$. Given $\varepsilon>0$, by uniform continuity of $f$ there exists $\delta>0$ such that for all $x,y\in X$, $d(x,y)<\delta$ implies $d(f(x),f(y))<\varepsilon$. Now this same $\delta$ works for the restriction $f\vert_A$, because if you have $x,y\in A$ with $d(x,y)<\delta$, then $x$ and $y$ are also in $X$, so $d(f(x),f(y))<\varepsilon$.

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    Sorry, I had deleted the comment accidentally. We were talking about the condition above for continuous extensions, and I attempted a sketch of a proof: Let $x \in \overline{A}$ but not $A$. Consider sequences in $A$ converging to $x$: these sequences are Cauchy, and since $f$ is uniformly continuous, evaluating each of these sequences termwise by $f$ are also Cauchy. They converge uniquely (interweave two, and we have convergent subsequences in a Hausdorff space) to a point $y$. Associating $f(x) = y$, and doing this for all limit points, gives a continuous extension to $\overline{A}$.2011-07-04
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In regards to your "too strong" conclusions there's a slightly more general statement one can make. Let $f: (X,d_1) \rightarrow (Y,d_2)$ be a continuous function and $A \subset X$ compact. Consider $g:=f|_A$ then $g$ is continuous (with A given the metric induced from $X$). Then we have that $g$ is a continuous function whose domain is compact, so $g$ is uniformly continuous. That is $f$ is uniformly continuous on $A$.