Consider the subset $Y$ below as a candidate for being a subspace of $\complex{2}$ \begin{equation*} Y=\setparts{\colvector{x_1\\x_2}}{x_1\in{\mathbb Z},\,x_2\in{\mathbb Z}} \end{equation*} ${\mathbb Z}$ is the set of integers, so we are only allowing “whole numbers” as the constituents of our vectors. Now, $\zerovector\in Y$, and additive closure also holds (can you prove these claims?). So we will have to try something different. Note that $\alpha = \frac{1}{2}\in\complex{\null}$ and $\colvector{2\\3}\in Y$, but \begin{equation*} \alpha\vect{x}=\frac{1}{2}\colvector{2\\3}=\colvector{1\\\frac{3}{2}}\not\in Y \end{equation*} So $Y$ fails the scalar multiplication closure requirement of either Property SC or Theorem TSS, and is therefore not a subspace.