Is the set $\{(x, y) : |y| = |x|\}$ a subspace of $\mathbb{R^2}$?

Question

So, I know that the set $H =\{(x, y) : |y| = |x|\}$ is not a subspace of $\mathbb{R^2}$, but I wanted to know if the following reasoning was correct:

$|0|=|0|$, and so the zero vector is in the set $H$.

If $(-x,y)$ $\in$ $H$ and $(x,y)$ $\in$ $H$, then $(-x,y)$ + $(x,y)$ = $(0,2y)$ $\notin$ $H$.

If $(x,y)$ $\in$ $H$, then $|x|=|y|$. Thus, $c|x| = c|y|$, and by the absolute value property of the reals, it follows that

$|c||x| = |c||y|$,

$|cx| = |cy|$.

Therefore, $(cx,cy)$ $\in$ $H$.

Since $H$ does not satisfy closure under addition, it is not a subspace of $\mathbb{R^2}$.

Answer 1

Your argument is right, but you don't need to say that many things. Your $H$ will fail to be a subspace as long as it fails one of the properties for a single choice of vectors.

So it is enough to say $(-1,1)\in H$, $(1,1)\in H$, $(0,1)=(-1,1)+(1,1)\not\in H$, so $H$ fails additivity and it is not a subspace.