Example NSC2A A non-subspace in $\complex{2}$, additive closure

Consider the subset $X$ below as a candidate for being a subspace of $\complex{2}$ \begin{equation*} X=\setparts{\colvector{x_1\\x_2}}{x_1x_2=0} \end{equation*}

You can check that $\zerovector\in X$, so the approach of the last example will not get us anywhere. However, notice that $\vect{x}=\colvector{1\\0}\in X$ and $\vect{y}=\colvector{0\\1}\in X$. Yet \begin{equation*} \vect{x}+\vect{y}=\colvector{1\\0}+\colvector{0\\1}=\colvector{1\\1}\not\in X \end{equation*}

So $X$ fails the additive closure requirement of either Property AC or Theorem TSS, and is therefore not a subspace.