# Mapping $\mathbb{R}^n$ "close" to integer points

For $$\epsilon>0$$ and a given $$n$$, define $$S_\epsilon=\bigcup_{x\in\mathbb{Z}^n}B_\epsilon(x)\subset\mathbb{R}^n$$ where $$B_\epsilon(x)$$ is the open ball of radius $$\epsilon$$ around the point $$x$$ in $$\mathbb{R}^n$$. Essentially, $$S_\epsilon$$ is the set of points in $$\mathbb{R}^n$$ that are "close enough" to (within $$\epsilon$$ of) an integer point. Define $$f:\mathbb{R}^n\times\mathbb{R}_{>0}\to\mathbb{R}$$ with $$f(x,\epsilon)=\inf\{\delta\geq1\mid\delta x\in S_\epsilon\}$$Note $$\delta x=(\delta x_1,\dots,\delta x_n)$$. Intuitively, we see this function is well defined as every point is "close enough" to a rational point, which can itself be multiplied to an integer. I have two main questions surrounding this function:

1) For what values of $$\epsilon$$ is this function continuous? Clearly, if $$\epsilon$$ is large enough, depending on the dimention, then $$f\equiv 1$$ everywhere. Can $$\epsilon$$ be made smaller and preserve continuity?

2) Is $$f$$ smooth, either with respect to $$x$$ or $$\epsilon$$? This one seems a little bit more hopeless, as the function essentially has "hills" which plateau on $$S_\epsilon$$. If we were looking at closed balls around integer points instead of open, would that change anything with respect to smoothness?

While those two questions are the main focus of this post, I would greatly appreciate any additional remarks about $$f$$. Thank you!

The answer to both question is no. The point $$(\frac{ \epsilon} {\sqrt n}, \dots, \frac{ \epsilon} { \sqrt n}, \epsilon)$$ for $$\epsilon<\frac{ \sqrt n} {2}$$ is a counterexample
• I guess this was definitely easier than I thought... Is there a continuous mapping on $\mathbb{R}^n} taking every point close to some integer point? Commented Mar 5, 2019 at 0:31 • I suppose by "close to some integer point" you mean inside some$S_\epsilon$for a fixed$\epsilon$. If$\epsilon\le\frac12$(replace this with a strict inequality if you use closed balls instead), the only continuous maps from$\mathbb{R}^n$to$S_\epsilon$are maps whose image lies entirely within one of the balls, since otherwise the image will be disconnected (but$\mathbb{R}^n$is connected). For$\epsilon>\frac12$, we may take a curve (say parametrized by$t$) starting at the origin that intersects every ball and map$x$to the point with$t=\|x\|\$. Commented Mar 5, 2019 at 20:17
1) Indeed, $$f(\cdot, \varepsilon)$$ is continuous iff $$\varepsilon$$ is big enough for the closure of $$\bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)$$ to cover completely $$\mathbb{R}^n$$. Assume otherwise. Translating, we can assume to have an open set $$O \subset B_1(0)$$, $$O \subset \mathbb{R}^n - \bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)$$.
Let us denote $$k_0 = (1,0,...,0)$$ and $$k = (\lceil \frac{1}{\varepsilon} \rceil,0,...,0)$$. Let $$x \in k+O$$. I am going to show that $$f$$ is not continuous somewhere between $$0$$ and $$x$$. Let $$y := tx$$ with $$t := \inf \{t \in [0,1]\ |\ \forall s \in [t,1], sx \notin \bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)\}$$. We have $$\big|\frac{1}{\lceil 1/\varepsilon \rceil} x - k_0\big| = \frac{1}{\lceil 1/\varepsilon\rceil}|x-k| < \varepsilon$$ because $$x-k \in O \subset B_1(0)$$. Hence $$\frac{1}{\lceil 1/\varepsilon\rceil}x \in B_{\varepsilon}(k_0)$$, so $$t \ge \frac{1}{\lceil 1/\varepsilon\rceil} > 0$$. Hence for some $$t'$$ in a neighborhood of $$t^-$$, we have $$f(t'x) = 1$$. And for $$t' \in [t,1]$$, for $$s \in [t', 1]$$, $$sx \notin \bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)$$ so $$f(t'x) = \frac{1}{t'} f(x) \ge\frac{1}{t'} > \frac{1}{t} \ge 1$$. Hence $$f$$ is not continuous at $$t$$.
2) Same as before, if $$\varepsilon_0$$ is the smallest value of $$\varepsilon$$ such that the closure of $$\bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)$$ is $$\mathbb{R}^n$$, $$f$$ is not continuous with respect to $$\varepsilon$$ on $$]0,\varepsilon_0$$. Consider $$y$$ as it was defined in the first part (not that it was on the boundary of $$\bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon}(z)$$). We had $$f(y,\varepsilon) = \frac{1}{t}f(x) > \frac{1}{t}$$. For any $$\varepsilon' > \varepsilon$$, $$y \in \bigcup \limits_{z \in \mathbb{Z}^n} B_{\varepsilon'}(z)$$ so $$f(y,\varepsilon')=1$$. Hence the diccontinuity at all $$\varepsilon < \varepsilon_0$$.