# Proving that $x^2$ is not uniformly continuous

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.

• 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. – Stephen Montgomery-Smith Aug 19 at 3:55
• I thought so as well, except I had difficulty proving that a set of that form is a counterexample. – ilovebulbasaur Aug 19 at 3:56

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 $$|f(x) - f(y) < \epsilon$$.
• 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 $|x-y|<\frac{1}{2}$. – ilovebulbasaur Aug 19 at 4:25
• I want to make sure that we don't have $x$ and $y$ in different intervals of the form $(n,n+\frac1{n^2})$. – Stephen Montgomery-Smith Aug 19 at 4:26