# Example that is not a subspace

I did some linear algebra exercise and did the following: Give an example of a nonempty subset $U$ of the xy-plane with the property that $U$ is closed with respect to scalar multiplication but $U$ is not a subspace.

I believe the set $U=\{(x,y): xy=0\}$ should provide such an example. Are there any better examples? More visual ones?

• Your example is perfect. Visually, it represents the union of the $x$ and $y$ axes (which is a union of two lines through the origin that are not collinear). It fails to be a subspace because it isn't closed with respect to vector addition. Aug 7, 2013 at 10:29
• @Adriano Sorry, I don't see how it is the union of the x and the y axis? Could you say a bit more?
– newb
Aug 7, 2013 at 10:37
• @newb: The points for which $xy=0$ are exactly the ones that are either on the $x$ axis or on the $y$ axis (or both). Aug 7, 2013 at 10:38
• @HenningMakholm Oh yes, right. Thank yoU!
– newb
Aug 7, 2013 at 10:40

Your example is perfect. Visually, it represents the union of the $x$ and $y$ axes (which is a union of two lines through the origin that are not collinear). It fails to be a subspace because it isn't closed with respect to vector addition.
To see your example is the union of the $x$ and $y$ axes, observe that: \begin{align*} U &= \{(x,y): xy=0\} \\ &= \{(x,y): x=0 \text{ or } y=0\} \\ &= \{(x,y): x=0\} \cup \{(x,y):y=0\} \\ &= (y\text{-axis}) \cup (x\text{-axis}) \end{align*}