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We know that $f(x)=x^2$ is not uniformly continuous as a function $f:\mathbb{R}\rightarrow[0,\infty)$. Indeed, let $\epsilon=1$. For any $\delta>0$, we may choose $\alpha>0$ large enough so that $\alpha\delta+\delta^2/4\geq \epsilon$. Then if we set $$x=\alpha$$ $$y=\alpha+\frac{\delta}{2}$$ we find $|x-y|<\delta$, yet $|f(x)-f(y)|\geq\epsilon$. Hence the $\epsilon-\delta$ definition of uniform continuity is negated and that $f$ is not uniformly continuous.

Now if $X\subset\mathbb{R}$ is any open unbounded set, how do we prove that $f:X\rightarrow [0,\infty)$ is not uniformly continuous? I tried following a similar procedure as above, but it didn't work out. The difficulty I am having is that I can't make sure that $y=\alpha+\delta/2\in X$, because $X$ could be an open unbounded set with narrower open intervals as $x$ increases, for example $$X=\bigcup_{n=1}^{\infty}(\sqrt{n},\sqrt{n}+\frac{1}{n}).$$

Given the above, is there a way to modify the above proof for the $f:X\rightarrow [0,\infty)$ case? I am not interested in just being given a proof, but I wanted to know how my proof might be modified, or if it just couldn't be modified in this case.

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  • $\begingroup$ I think what you are trying to prove isn't true. $\bigcup_n (n,n+\frac1{n^2})$ might be an easier example to consider. $\endgroup$ – Stephen Montgomery-Smith Aug 19 at 3:55
  • $\begingroup$ I thought so as well, except I had difficulty proving that a set of that form is a counterexample. $\endgroup$ – ilovebulbasaur Aug 19 at 3:56
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It is not true. Consider $X = \bigcup_n (n,n+\tfrac1{n^2})$. Note if $x,y \in (n,n+\tfrac1{n^2})$, then $$ |f(x) - f(y)| \le |f(n+\tfrac1{n^2}) - f(n)| = \tfrac2n + \tfrac1{n^2} \le \tfrac3n .$$ Given $\epsilon > 0$, choose $N > \frac3\epsilon$. If $x,y \in \bigcup_{n\ge N} (n,\frac1{n^2})$, and $|x-y| < \tfrac12$, then $|f(x) - f(y)| < \epsilon$. And since $f(x)$ is uniformly continuous on $[0,N+1]$, we can find $\delta > 0$ and $\delta < \tfrac12$ such that if $x,y \in [0,N+1]$, then $|x-y| < \delta$ implies $|f(x) - f(y) < \epsilon$.

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  • $\begingroup$ Thanks! One minor thing though: what does the $|x-y|<\frac{1}{2}$ do? Because from the choice of $N$, I thought that $x,y\in \bigcup_{n\geq N}(n,n+\frac{1}{n^2})$ is enough to guarantee $|f(x)-f(y)|<\epsilon$, so we didn't need to include $|x-y|<\frac{1}{2}$. $\endgroup$ – ilovebulbasaur Aug 19 at 4:25
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    $\begingroup$ I want to make sure that we don't have $x$ and $y$ in different intervals of the form $(n,n+\frac1{n^2})$. $\endgroup$ – Stephen Montgomery-Smith Aug 19 at 4:26

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