Richert's Theorem

Theorem
Let $S = \set {s_1, s_2, \dots}$ be an infinite set of (strictly) positive integers, with the property:
 * $s_n < s_{n + 1}$ for every $n \in \N$

Suppose there exists some integers $N, k$ such that every integer $n$ with $N < n \le N + s_{k + 1}$:
 * $n$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_k}$
 * $s_{i + 1} \le 2 s_i$ for every $i \ge k$

Then for any $n > N$, $n$ can be expressed as a sum of distinct elements in $S$.

Proof
We prove this using First Principle of Mathematical Induction.

Let $\map P n$ be the proposition:


 * For every integer $m$ with $N < m \le N + s_{n + 1}$:
 * $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_n}$.

Basis for the induction
From our assumption above, $\map P k$ is true.

This is the basis for the induction.

Induction Hypothesis
Suppose for some $n > k$, $\map P {n - 1}$ is true.

That is, for every integer $m$ with $N < m \le N + s_n$:
 * $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.

This is the induction hypothesis.

We need to show that $\map P n$ is true.

That is, for every integer $m$ with $N < m \le N + s_{n + 1}$:
 * $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_n}$.

Induction Step
This is the induction step:

By the induction hypothesis, for every integer $m$ with $N < m \le N + s_n$:
 * $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.

Hence we only need to consider $N + s_n < m \le N + s_{n + 1}$.

For $m$ in this range:
 * $N < m - s_n \le N + s_{n + 1} - s_n \le N + s_n$

By the induction hypothesis, $m - s_n$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.

Hence $m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_{n - 1}, s_n}$.

Hence $\map P n$ is true.

By the First Principle of Mathematical Induction, $\map P n$ is true for all $n \ge k$.

Since $1 \le s_n < s_{n + 1}$ for every $n \in N$:
 * $s_n \ge n$ for every $n \in \N$

Let $K > N$.

If $K - N \le k + 1$:
 * $N < K = N + \paren {K - N} \le N + k + 1 \le N + s_{k + 1}$

From our assumption $K$ can be expressed as the sum of distinct elements in $\set {s_1, s_2, \dots, s_{k + 1}}$.

If $K - N > k + 1$:
 * $N < K = N + \paren {K - N} \le N + s_{K - N}$

By the result above, since $K - N - 1 > k + 1 - 1 = k$:
 * $K$ can be expressed as the sum of distinct elements in $\set {s_1, s_2, \dots, s_{K - N - 1}}$.

Thus every number greater than $N$ can be expressed as the sum of distinct elements in $S$.