Please let me know how to approach this problem.
Show that if $U ≠ \{0\}$ is a subspace of $\mathbb{R}$, then $U = \mathbb{R}$
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linear-algebra
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2Presumably you are considering the real numbers as a real vector space (i.e., the scalars are real numbers). Is that correct? – 2011-07-11
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1Yup, I'm thinking about proving U -> R and R -> U so U = R, but what I don't get is the part given that U ≠ {0} – 2011-07-11
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0Larry, what do you mean by "proving U -> R and R -> U"? – 2011-07-11
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1like if U≠{0} is a subspace of R then U ⊆ R, and R ⊆ U as well so U = R. I don't know. I'm just guessing – 2011-07-11
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0Oh, you mean $U\subseteq \mathbf R$ and $\mathbf R\subseteq U$? Yeah, $U\subseteq \mathbf R$ is part of the definition of subspace. The other containment is the point, and $U\neq\{0\}$ is required because $\{0\}$ is a subspace that is not equal to $\mathbf{R}$. – 2011-07-11
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0@Larry I have given a detailed step-by-step answer and 5 exercises (which might be useful to you since they are very pertinent to your question). – 2011-07-11