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Could anyone provide an example of a nonempty subset $U$ of $R^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $R^2$

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5 Answers 5

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The cross.

(More precisely, the union of the $x$-axis and $y$-axis.)

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  • $\begingroup$ +1: Your answer inspired me to do the opposite (not exactly, but near enough). $\endgroup$
    – copper.hat
    Jan 11, 2016 at 3:44
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Take the set of all points sitting on either the x or y axis, i.e. $\left\{(x, y) | x = 0 \mbox{ or } y = 0\right\}$. This is clearly a subset of $\mathbb{R}^2$ that is closed under scalar multiplication (because $(x, 0) \times c = (cx, 0)$ and similarly for $(0, y)$), but it is not closed under addition (because $(1, 0) + (0, 1) = (1, 1)$ is not in the set), and hence it is not a subspace.

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Take $\mathbb{R}^2 \setminus (\{(0,y) | y \neq 0 \} \cup \{(x,0) | x \neq 0 \})$.

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    $\begingroup$ +1 for the "atheistic" approach : P. $\endgroup$
    – Aloizio Macedo
    Jan 11, 2016 at 3:47
  • $\begingroup$ @AloizioMacedo: :-). $\endgroup$
    – copper.hat
    Jan 11, 2016 at 3:47
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The complement of a 1-Dimensional subspace (together with {(0,0)}).

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Consider $U= \{ (x_{1},x_{2}) | x_{1} x_{2} =0 \text{ and }x_{1},x_{2} \in \mathbb{R}\}$

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