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Find an example of a nonempty subset $U$ of $\mathbb{R}^2$ where $U$ is closed under scalar multiplication but U is not a subspace of $\mathbb{R}^2$. I saw other questions asking the same and some of the answers were that any two lines through the origin work. But for my example I'm not sure if I'm doing it correct:

My example: Let $U = \left\{(x,y) : y = x \right\} \cup \left\{ (x, y) : y = -x\right\}$

To show this is not closed under vector addition let $a = (x_1 , x_2)$ and $b = (y_1, y_2)$ such that $x_2 = x_1$, $x_2 = -x_1$ and similar results for $y$.

Then $a + b = (x_1 + y_1, x_2 + y_2)$ such that $x_1 + y_1 = x_2 + y_2$ and $x_1 + y_1 = -x_2 - y_2$.

This seems closed under addition since all those values are $0$. Am I doing this completely wrong?

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  • $\begingroup$ Best not to use arbitrary vectors and give a concrete example instead $a=(1,1)$ and $b=(2,-2).$ Then $a+b=(3,-1).$ $\endgroup$
    – Melody
    Nov 17, 2018 at 19:05

3 Answers 3

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Your counterexample should be two specific vectors in $U$ (e.g., $(1,1),(1-1)$) instead of arbitrary vectors (i.e. $(x_1,x_2),(y_1,y_2)$).

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    $\begingroup$ So my example is correct as long as I just use specific values? $\endgroup$ Nov 17, 2018 at 19:07
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    $\begingroup$ Yeah, you're correct. In general, to disprove a statement you just have to find a single counterexample. $\endgroup$
    – ATOMP
    Nov 17, 2018 at 19:08
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$(1,1) + (2,-2) = (3,-1)$ which is not in $U$ though both $(1,1)$ and $(2,-2)$ are.

One example is enough to disprove closedness under sums.

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The union of $x$ axis and $y$ axis is not a subspace but it is closed under scalar multiplication.

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