partition of the set of natural numbers

Question

Let $S=\{s_i\}_{i\geq 0}$ any infinite set of positive integers for which $c=\max \{|s_{i+1}-s_i|:i \geq 0\}$ exists.

Let $T_0=S$ and define the sets $T_1, T_2,\dots, T_{c-1}$ recursively by $$T_j=\{s+j: s\in S\}-\bigcup \limits_{i=0}^{j-1}T_i,$$ i.e., just translate $S$ by $j$, and make sure that $T_j$ is disjoint from previous $T_i$'s. Since $|s_{i+1}-s_i|\leq c$ for all $i\geq 0$, the sets $T_0, T_1, T_2,\dots, T_{c-1}$ provide a partition of $\mathbb{Z}^+$.

I am not able to grasp some moments of above excerpt. So let me ask some questions:

1) Why the author considers only $T_0, T_1, T_2,\dots, T_{c-1}$? What if we add also $T_c, T_{c+1}, T_{c+2}$?

2) Also I cannot able to show that $T_0\sqcup T_1\sqcup T_2\sqcup \dots \sqcup T_{c-1}=\mathbb{Z}^+$. I have problems with showing that $\mathbb{Z}^+ \subset T_0\sqcup T_1\sqcup T_2\sqcup \dots \sqcup T_{c-1}$.

Suppose $m\in \mathbb{Z}^+$ then $s_k\leq m <s_{k+1}$. If $m=s_k$ then $m\in T_0$ so assume that $s_k< m <s_{k+1}$ then $0<m-s_k<s_{k+1}-s_k \leq c$ so $m-s_k=t$ where $1\leq t \leq c-1$. Thus, $m\in T_0+t$. But how to show that $m\notin T_0\cup \dots T_{t-1}$?

Can anyone explain my questions, please.

Answer 1

First of all, your claim is false because $T_0 \cup \ldots \cup T_{c-1}$ is a partition of $\mathbb{Z} \cap [\min S, \infty)$, in particular if $\min S = 5$, then $\mathbb{Z}^+ \ni 4 \notin (T_0\cup \ldots \cup T_{c-1})$.

We can fix this easily assuming that $\min \mathbb{Z}^+ \in S$.

To answer your first question, we do not consider $T_c$ and other such sets, because they are empty which is because $T_0 \cup \ldots \cup T_{c-1}$ already cover whole $\mathbb{Z}^+$.

Your second question can be solved in two ways:

1. Observe that if $m \in T_l$ for $l < t$, then you are also done, because even if $m \notin T_t$, then it is in $T_l$, that is, it does belong to one of these sets, and so belongs to their union.
2. Instead of picking any $s_k \in S$ pick $s_k = \max (S \cap [\min \mathbb{Z}^+,m])$, this way it is the biggest number smaller than $m$, and having $m \in T_l$ for $l < t$ would contradict your choice of $s_k$.

I hope this helps $\ddot\smile$