Sure -- take the union of \\(\operatorname{\uparrow}(\\{1\\})\\) and \\(\operatorname{\uparrow}(\\{3\\})\\). This is an upset because anything above an element in the union is above an element in one of the individual upsets, hence is itself in one of the upsets, and therefore is in the union. If this were a principal upset, \\(\\{1\\} \wedge \\{3\\} = \emptyset\\) would need to be in the upset (since principality implies a _single_ minimal element). But it isn't, so it's not.

**Puzzle JMC1:** Show that a principal upset must have a unique minimal element.

**Puzzle JMC2:** Show that if \\(x, y \in \operatorname{\uparrow}(z)\\) and \\(x \wedge y\\) exists, then \\(x \wedge y \in \operatorname{\uparrow}(z)\\).

**Puzzle JMC1:** Show that a principal upset must have a unique minimal element.

**Puzzle JMC2:** Show that if \\(x, y \in \operatorname{\uparrow}(z)\\) and \\(x \wedge y\\) exists, then \\(x \wedge y \in \operatorname{\uparrow}(z)\\).