# Determine if a set is a subspace or not

I'm having trouble understanding the conditions of the set $$H = \{(x,y) \in \Bbb R^{2} \mid\text{ either } x,y \ge0 \text{ or }x,y \le0 \}.$$

To determine if it's closed under addition, let $u$ be $x_1, y_2$ and $v$ be $y_1, y_2$. It is closed if $u + v$ exists. I take $x_1 + x_2$ and $y_1 + y_2$, but I am unsure how to explain because of the conditions.

To determine if it's closed under scalar multiplication, I let $C$ be a negative scalar and $u$ be vector $x,y$. I was thinking if I choose a negative scalar, and choose the option that $x_1, y_2$ would have to be larger than $0$, then it would not be closed under scalar multiplication and thus not a subspace.

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You omitted the parentheses around the ordered pairs; they’re required, so you should edit the question to include them. – Brian M. Scott Nov 2 '12 at 19:25
As a matter of fact, $(0,0)\notin H$, but every subspace contains $(0,0)$. – Hagen von Eitzen Nov 2 '12 at 19:37
Why $\,(0,0)\notin H\,$? In fact it is. – DonAntonio Nov 3 '12 at 11:12

Your $H$ is closed under multiplication by scalars, but it is not closed under addition. If you consider $(2,1)$, $(-1,-2)$, then both are in $H$ but $(2,1)+(-1,-2)=(1,-1)$ is not in $H$.
$(1,3)+(-3,-2)=(-2,1)$ so $H$ is not closed under addition and therefore is not a subspace of $\mathbb R^2$. (To help you visualise the set, it consists of all elements in the first and third quadrants on the Cartesian plane, including the axes.)
The conditions are essentially that $(x,y)\in H$ if $(x,y)$ is in the first or third quadrant. To demonstrate that $H$ is not closed under addition, just show that the sum of some two vectors in the first and third quadrant is in the second or fourth quadrant.