A proof I'm reading begins as follows:
Let $x_1$ be an internal point of $K_1$ and $x_2$ any point of $K_2$ (here $K_1$ and $K_2$ are nonempty disjoint convex subsets of a linear space $X$). Define: $$z=x_2-x_1 \qquad K=K_1+[-K_2]+z$$ Then $K$ is a convex set that contains the origin as an internal point and does not contain $z$
I'm having trobule understanding why $z \notin K$. In particular, I see $K$ has been shifted by $z$ so that what was $-z \in K_1+[-K_2]$ is now the origin in $K=K_1+[-K_2]+z$ -- so in particular not $z$. But why can't some other other point in $K_1+[-K_2]$ get shifted to $z \in K$?
If $z \in K$, then $z=k_1-k_2+z$ for some $k_1 \in K_1$ and $k_2 \in K_2$. This means $k_1=k_2$. However $K_1 \cap K_2=\emptyset$, therefore $z \not\in K$.