# Does span(A) < span(B) imply A < B?

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 $$A \subseteq B$$?

## 2 Answers

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, $$A \not\subset B$$.

• 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. Jan 21, 2020 at 5:28
• Yes, true. I forgot that here, thank you for pointing it out. Jan 21, 2020 at 13:26

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 $$(1,0)\notin B$$.