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This question arose while reading C.W. Edwards, Jr.'s Advanced Calculus of Several Variables, Section IV-2, proof of Theorem 2.2. I am fairly sure the functions describe below are uniformly continuous, but that is contrary to my long-held, intuitive concept of continuity.

This is the definition of uniform continuity I am following: for some domain $\mathscr{D}\subset\mathbb{R}$ the function $f:\mathscr{D}\to\mathbb{R}$ is uniformly continuous on $\mathscr{D}$ if and only if, given $\epsilon>0$ there exists $\delta>0$ such that $x,y\in\mathscr{D}$, $\left|x-y\right|<\delta\implies\left|f\left[x\right]-f\left[y\right]\right|<\epsilon$.

Now, define $\Delta\mathbb{N}=\{x\in\mathbb{R}\vert\exists n\in\mathbb{N}\wedge\left|x-n\right|<\Delta\}$, where $0<\Delta\ll1/2$. That is, those points of $\mathbb{R}$ lying in some open $\Delta$-neighborhood of some integer $n$.

I believe $f[x]=\left\lfloor x\right\rfloor$ is uniformly continuous on the closed set $\mathcal{D}_{\Delta}=\{x\vert x\in\mathbb{R}-\Delta\mathbb{N}\}$. Because given $\epsilon>0$ there exists $\delta>0$ such that $\left|x-y\right|<\delta\implies\left\lfloor x\right\rfloor =\left\lfloor y\right\rfloor$ for some $x,y\in\mathcal{D}_{\Delta}$. In fact, $\left|x-y\right|<\delta\le\Delta\implies\left\lfloor x\right\rfloor =\left\lfloor y\right\rfloor$ .

I also believe that $f[x]=\left\lfloor x\right\rfloor$ is uniformly continuous on the open set $\mathcal{D}=\{x\vert x\in\mathbb{R}-\mathbb{N}\}$. This is true because every point $x\in\mathcal{D}$ lies in some open interval $\left(x-\delta,x-\delta\right)\subset\left(n-1,n\right)$ where $n-1<x<n$ and $\delta$ is sufficiently small. So for $x\in\mathcal{D}$ there exists $y\in\left(x-\delta,x-\delta\right)$ such that $\left\lfloor x\right\rfloor =\left\lfloor y\right\rfloor$.

Furthermore; I believe the identity function $\mathcal{I}[x]=x$ is uniformly continuous on both $\mathcal{D}_{\Delta}$ and $\mathcal{D}$. The same approach as used above applies to this case.

Am I correct in these statements?

Steven Thomas Hatton
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  • You're correct that $x\mapsto \lvert x\rvert$ is uniformly continuous on $\mathbb{R}\setminus \Delta\mathbb{N}$, but that mapping isn't uniformly continuous on $\mathbb{R}\setminus \mathbb{N}$ because your $\delta$ will be dependent on $x$. – Michael L. Aug 30 '17 at 19:36
  • The identity is uniformly continuous on any set in a trivial sense. For any $\epsilon > 0$, let $\delta = \epsilon$. Then $\lvert x-y\rvert < \delta \Rightarrow \lvert x-y\rvert < \epsilon$. – Michael L. Aug 30 '17 at 19:38
  • So, would I be correct in saying that $\left\lfloor x\right\rfloor$ is continuous on $\mathbb{R}\backslash\mathbb{N}$, but not uniformly continuous? – Steven Thomas Hatton Aug 30 '17 at 19:57
  • Right. The logic you used is fine for proving continuity, but to prove uniform continuity, you have to show that the $\delta$ you choose is dependent only on $\epsilon$, not on $x$. – Michael L. Aug 30 '17 at 19:58

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$x \mapsto \lfloor x \rfloor$ is not uniformly continuous on $\mathbb R - \mathbb N$. No matter how small your $\delta$ is, you can choose $x$ and $y$ so that $|x - y| < \delta$ and yet $\lfloor x \rfloor \ne \lfloor y \rfloor$. For example, take $x = -\delta / 3$, $y = \delta / 3$.

Ayman Hourieh
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