Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $R^2$

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer

The cross.

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

share|cite|improve this answer
+1: Your answer inspired me to do the opposite (not exactly, but near enough). – copper.hat Jan 11 at 3:44

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

share|cite|improve this answer
+1 for the "atheistic" approach : P. – Aloizio Macedo Jan 11 at 3:47
@AloizioMacedo: :-). – copper.hat Jan 11 at 3:47

The complement of a 1-Dimensional subspace (together with {(0,0)}).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.