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Given $A \subseteq B \subseteq V$ for some vector space $V$, we know that $\mathrm{span}(A) \subseteq \operatorname{span}(B) $. We can see that as span of $B$ can make any elements of $A$.

However does the inverse of the statement hold? So if $\operatorname{span}(A) \subseteq \operatorname{span}(B) $, can we say $A \subseteq B$?

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2 Answers 2

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No. For example, take $V = \mathbb{R}^2$, $A = \{(2,0)\}$ and $B = \{(1,0), (0,1)\}$. Since $B$ is a basis for $V$, $\operatorname{span}(B) = V$, so $\operatorname{span}(A) \subseteq \operatorname{span}(B)$. However, $A \not\subset B$.

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    $\begingroup$ If you write \mathrm{span} rather than \operatorname{span} then instead of $\operatorname{span}A$ you'll see $\mathrm{span}A,$ without proper spacing. That doesn't mean just that \operatorname{} adds spaces to its left and right, but rather that it has context-dependent spacing. $\endgroup$ Jan 21, 2020 at 5:28
  • $\begingroup$ Yes, true. I forgot that here, thank you for pointing it out. $\endgroup$
    – Vincent
    Jan 21, 2020 at 13:26
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No. In $\mathbb{R}^2$, take $A=\{(1,0)\}$, $B=\{(2,0),(0,1)\}$. Then $\mathrm{Span}\, A=\mathbb{R}\subseteq\mathbb{R}^2$ and $\mathrm{Span}\, B=\mathbb{R}^2$ (so $\mathrm{Span}\, A\subseteq\mathrm{Span}\, B$), but $A\not\subseteq B$, since $(1,0)\notin B$.

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