Show that, if $S_{1}\subseteq S_{2}$, then $\text{span}(S_{1})\subseteq\text{span}(S_{2})$.

Question

Show that if $S_{1}$ and $S_{2}$ are subsets of a vector space $V$ such that $S_{1}\subseteq S_{2}$, then $\text{span}(S_{1})\subseteq\text{span}(S_{2})$. In particular, if $S_{1}\subseteq S_{2}$ and $\text{span}(S_{1}) = V$, deduce that $\text{span}(S_{2}) = V$.

MY ATTEMPT

As to the first part, let's suppose that $s\in\text{span}(S_{1})$, where $s = a_{1}s_{1} + as_{2} + \ldots + a_{n}s_{n}$ for some $s_{i}\in S_{1}$ and $a_{i}\in\textbf{F}$.

Since $s_{i}\in S_{2}$, we conclude that $a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n}\in\text{span}(S_{2})$, that is to say, $\text{span}(S_{1})\subseteq\text{span}(S_{2})$.

As to the second part, we have \begin{align*} S_{1}\subseteq S_{2}\subseteq V \Longrightarrow \text{span}(S_{1})\subseteq\text{span}(S_{2})\subseteq\text{span}(V) \Longrightarrow V\subseteq\text{span}(S_{2})\subseteq V \end{align*} from whence we conclude that $\text{span}(S_{2}) = V$, as desired.

Can someone double-check my reasoning?

Answer 1

This can be done using only the universal properties of the span (that is, its “top-down definition”, as opposed to its bottom-up definition/description as the collection of all linear combinations of elements of the set).

Namely, $\mathrm{span}(X)$ is the unique subspace of $V$ satisfying:

1. $X\subseteq \mathrm{span}(X)$; and
2. If $W$ is a subspace of $V$ such that $X\subseteq W$, then $\mathrm{span}(X)\subseteq W$.

So: we know that $S_1\subseteq S_2\subseteq\mathrm{span}(S_2)$. Since $\mathrm{span}(S_2)$ is a subspace that contains $S_1$, it follows that $\mathrm{span}(S_1)\subseteq \mathrm{span}(S_2)$.

Answer 2

Let $S_1=\{s_{1,\alpha}\}_{\alpha\in I}$ and $S_2=\{s_{2,\beta}\}_{\beta\in J}$ with $S_1\subset S_2$. Let's show that $\langle S_1\rangle \subset \langle S_2\rangle $. Take any $x\in \langle S_1\rangle $ then $x=\sum \limits_{\alpha\in I}\lambda_{\alpha} s_{1,\alpha}$ where $\lambda_{\alpha}\neq 0$ for finitely many $\alpha$'s and since $s_{1,\alpha}\in S_1\subset S_2$ then it follows that $x\in \langle S_2\rangle $. Which means that $\langle S_1\rangle \subset \langle S_2\rangle $.

Answer 3

Alternatively, for $S \subseteq V$ define $$\mathcal{F}_S := \{ H \subseteq V :\, H \textrm{ is a subspace of $V$ and $S \subseteq H$} \}.$$ With this, we have $\operatorname{span}(S) = \bigcap \mathcal{F}_S$.

So, if $S_1 \subseteq S_2$, then $\mathcal{F}_{S_2} \subseteq \mathcal{F}_{S_1}$ and then $\bigcap \mathcal{F}_{S_1} \subseteq \bigcap \mathcal{F}_{S_2}$. Observe that this proof doesn't require to consider the case when one subset is the empty set, and yours does.