I'm given the question
Draw the Hasse diagram of the partially-ordered set of of connected subgraphs, up to isomorphism, of the graph $K_{2,1}$ ordered by containment. Show that if the qualifier "up to isomorphism" is removed, then the resulting partially ordered set is not a lattice."
I understand the first part well enough. I just create a poset of the 5 non-isomorphic subgraphs and then draw the Hasse diagram with an $\{0\}$ at the bottom and the graph $K_{2,1}$ at the top.
I'm confused about the second part however. Wouldn't any pair of subgraphs of $K_{2,1}$ in a Hasse diagram share the upperbound $K_{2,1}$ and lower bound $\{0\}$?