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Show that, if $S_1$ and $S_2$ are subsets of a vector space $V$ such that $S_1 \subseteq S_2$, then $span(S_1) \subseteq span(S_2)$.

A (hypothesis): $S_1$ and $S_2$ are subsets of a vector space $V$ such that $S_1 \subseteq S_2$.

B (conclusion): $span(S_1) \subseteq span(S_2)$

A1: Let $S_2 = \{x^n, x^{n - 1}, ..., x^0\}$

A2: $span(S_2) = \{a_nx^n + a_{n - 1}x^{n - 1}+ ... + a_0 : a \in F \}$

A3: Let $S_1 = \{w^n, w^{n - 1}, ..., w^0 \}$ where $w = a_nx^n + a_{n - 1}x^{n - 1}+ ... + a_0$.

$\therefore$ The elements of $S_1$ are linear combinations of $S_2$.

A4: $span(S_1) \subseteq span(S_2)$

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness. If there are any errors, then please explain why and what the correct solution is.

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    The question didn't say that elements of $S_1$ were linear combinations of elements of $S_2$. It said elements of $S_1$ _are also elements of_ $S_2$. That's what $S_1\subseteq S_2$ means. It simplifies the proof somewhat. – 2017-02-14
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    @ziggurism thanks for the response. But how is that different from what I've done? The elements of $S_1$ would still be equal to linear combinations of the elements in $S_2$? Or have I made an error? – 2017-02-14
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    a) you assume without justification that $S_1,S_2$ are finite. b) You do not actually show something and simply write $\therefore$ (though admittedly there is little to show). - In order to make the proof more formal: Can you express "$v$ is an element of $Span(S)$" without the use of "$\ldots$"? - Alternatively, do you know that $Span(S)$ can alo be characterized as the smallest vector subspace of $V$ that contains $S$? – 2017-02-14
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    My complaint isn't that you made an error, but rather that you made your argument less simple than it could be. In step A3, instead of assuming elements of $S_1$ are linear combos, simply assume they are some vectors in $S_2$. Also, do not assume there are the same number $n$ of them, that actually is an error. – 2017-02-14
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    @ziggurism I see what you're saying. What do you mean by not assuming there are the same number $n$ of them? Can you please show me what you mean? – 2017-02-14
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    @HagenvonEitzen thanks for the response. How would I not use "..."? My textbook uses these frequently? – 2017-02-14

2 Answers 2

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It's right, if I were you I'd elaborate a bit more on the passage from A3 to the conclusion though

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    Thanks for the response. What do you suggest for elaboration? – 2017-02-14
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    First of all, since $S_1 \subset S_2$ one easier way to prove it is the following: Span($S_1$) is the intercession of all finite linear combinations of elements of subsets of $S_1$. In particular since $S_1 \subset S_2$ you get Span($S_1$) $\subset$ Span($S_2$) – 2017-02-14
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what you are proving is $S_1\subset span(S_2)\implies span(S_1)\subset span(S_2)$ but what you need is even simpler.

Let $S_2=\lbrace x_1,...,x_n\rbrace$ and without loss of generality $S_1=\lbrace x_1,...,x_k\rbrace$, with $k\leq n$ since $S_1\subset S_2$ (Let guess $k

Let $w\in span(S_1)$, so exists $a_1,...,a_k\in F$ s.t. $w=\sum_{i=1}^ka_ix_i$. Taking $a_{k+1}=...=a_n=0\in F$, $w=\sum_{i=1}^na_ix_i\in span(S_2)$