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Problem 39 (The Span Of A Set Of Integers Is Closed)

Let $S$ be a nonempty set of integers. Prove that $\text{span}(S)$ is closed under integer linear combinations.

Solution

In order to show that $\span(S)$ is closed, we will show that $\span(S) = \span(\span(S))$. We will first prove that $\span(\span(S))\subseteq \span(S)$. Let $\sigma_1 \in \span(\span(S))$. This means that we can write it as $\sigma_1 = c_1a_1 + c_2a_2 + \cdot\cdot\cdot + c_na_n$ such that $a_i \in \span(S)$, and $c_i \in \mathbb{Z}$. Because $a_i \in \span(s)$, we know that we can write each $a_i$ as an integer linear combination of elements in $S$. Since every $a_i \in \span(S)$, we know that every $a_i \in S$. Thus, we know that $\sigma_1$ is an integer linear combination of elements in $S$. Thus, we know that $\sigma_1 \in \span(S)$,

We will now prove that $\span(S)\subseteq \span(\span(S))$. Let $\sigma_2 \in \span(S)$. To be in $\span(\span(S))$, we know that $\sigma_2$ must be an integer linear combination of elements in $\span(S)$. We compute $\sigma_2 = 1 \cdot \sigma_2$. Thus, $\sigma_2 \in \span(\span(S))$. These two facts show that $\span(S) = \span(\span(S))$. Thus, $\span(S)$ is closed.

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