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Assuming that a function f is uniformly continuous, and starting from the ϵ-δ definition of continuity, how does one prove that it is also continuous on the real numbers?

Thanks.

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closed as off-topic by zarathustra, Jeel Shah, Etienne, Swapnil Tripathi, Mark Bennet Jan 13 '15 at 21:48

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    $\begingroup$ If you use the epsilon delta definition, there is only one answer: it is obvious. $\endgroup$ – Mister Benjamin Dover Jan 13 '15 at 20:15
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    $\begingroup$ Just cover up the "uniformly" part. $\endgroup$ – Alexander Gruber Jan 13 '15 at 20:24
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    $\begingroup$ This question appears to be off-topic because it is too obvious to be a question $\endgroup$ – Etienne Jan 13 '15 at 20:56
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    $\begingroup$ @Laters wrong claims, yeah sure. Not even worth to discuss. $\endgroup$ – Aaron Maroja Jan 13 '15 at 23:22
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    $\begingroup$ This is the most baffling comment chain I have ever read on this site. $\endgroup$ – Eric Stucky Jan 14 '15 at 12:04
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Let $f: X \to \mathbb{R}$ be uniformly continuous. Then for any $\epsilon > 0$ given there exists $\delta > 0$ such that $x,y \in X$,

$$|x - y| < \delta \Rightarrow |f(x) - f(y)|<\epsilon$$

In particular fix $a \in X$ arbitraly, then

$$|x - a| < \delta \implies |f(x) - f(a)| < \epsilon $$

for every $a \in X$ f is continuous.

An example in which $f$ is continuous but not uniformly continuous is $f(x) = \frac{1}{x}$, where $X = \mathbb{R}^+$ can you show why?

Added: Here is a hint to start. Let $0 <\epsilon<1 $, then for every $\delta > 0$ chosen we may take a natural number $n > \frac{1}{\delta}$ and set $x = \frac{1}{n}$ and $y = \frac{1}{2n}$. What is the contradiction?

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  • $\begingroup$ Feel free to ask. $\endgroup$ – Aaron Maroja Jan 13 '15 at 20:41
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Uniform continuity (of $f(x)$ on $A$): For every $\epsilon>0$ there is $\delta$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.

while

Continuity (of $f(x)$ at $a\in A$): For every $\epsilon>0$ there is $\delta$ such that if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

Given $\epsilon>0$ then, by uniform continuity, there is $\delta$ such that for all $x,y\in A$ such that $|x-y|<\delta$ implies that $|f(x)-f(y)|<\epsilon$. In particular if we take $y=a$ we get that for all $x\in A$ such that $|x-a|<\delta$ then necessarily $|f(x)-f(a)|<\epsilon$. This last is the definition of continuity.

The key observation is that the definition of uniform continuity becomes the definition of continuity when you look to a particular point $y=a$.

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