Same length open intervals cover $\mathbb{Z}$. Give some intervals: $D_1 = \bigcup_{i\in\mathbb{Z}}(a_i,b_i), D_2 = \bigcup_{i\in\mathbb{Z}}[b_i,a_{i+1}]$,
which like $\cdots,(a_{-1},b_{-1}),[b_{-1},a_0],(a_0,b_0),[b_0,a_1],(a_1,b_1), [b_1,a_2],\cdots$,
and $|(a_i,b_i)| = ks, |[b_i,a_{i+1}]| = s, \forall i \in \mathbb{Z}$, here $s \in \mathbb{R}, k \in \mathbb{Z}$. Now if give $a_0$, maybe some $s$ can let $D_1$ cover $\mathbb{Z}$.
My question is: if $s \neq 0$, does there exist $\min(s)$ (or $\inf(s)$), such that $D_1$ can cover $\mathbb{Z}$ for some choised $a_0$?
I am not sure I made the question clear, it is from my other question: If $\cos(nx+\phi) < \frac{\sqrt{3}}{2} \forall n \in \mathbb{Z}^+\cup\{0\}$, does there exist $\min(x)$?
 A: $s\ge 1$ means any $[b_i, a_{i+1}]$ contains an integer, so it can't be in $D_1$.
If $s <1, s \in \mathbb Q$, then that can be done. Setting $s=\frac{p}q,\, p,q > 0, p,q \in \mathbb Z$, then $a_0=\frac{s-1}2, k=q-1$ works, as that results in
$$a_i=\frac{s-1}2 + ip,\, b_i=-\frac{s+1}2 + (i+1)p,$$
and we get
$$ (i+1)p -1 < b_i < a_{i+1} < (i+1)p, \label{eq1} \tag{1}$$
using $\frac{s+1}2 <1$ and $\frac{s-1}2 < 0$, both simply derived from $s < 1$. \eqref{eq1} means that no $[b_i,a_{i+1}]$ can contain an integer, so $D_1$ covers $\mathbb Z$.
If $s <1$ is irrational, this can't be done. That's because all the $a_i=a_0+i(k+1)s$ will then have different fractional parts (which is $x-\lfloor x \rfloor$). Otherwise, $s$ would be a solution of a linear equation where two $a_0$'s are cancelling each other and the remaining parts are all integers, so $s$ would be rational).
That would mean the fractional parts of $a_i$ are at some point dense in $[0,1]$, so there exists different indices $u,v$ and an integer $h$ such that $a_u + h < a_v < a_u + h + s$. But that means
$$h < a_v-a_u = (v-u)(k+1)s < h +s.$$
$(v-u)(k+1)s$ is the difference of any 2 elements with index difference $v-u$.
That means the fractional parts of the sequence elements $a_0, a_{v-u}, a_{2(v-u)},\ldots$ increase by a positive amount, but less than $s$. So for one index $w$, the fractional part of $a_w$ will fall into the interval $(0,s)$, which means that $[b_{w-1}, a_w]$ will contain an integer, which means that integer is not in $D_1$.
To sum up: What you want is possible only for rational $s <1$, so the infimum is 0, which is however not attained.
