I came across a question as follows: Show that every asymmetric relation over a set $A$ is irreflexive.
The solution instructs one to use the relation < and suppose that it is asymmetric but not irreflexive, so that one derives a contradiction.
Since the relation is not irreflexive, there is an $a\in A$ with $a < a$. By asymmetry, not $a < a$ - contradiction"
But my understanding is that being non-irreflexive doesn't necessary mean it is reflexive, which seems to have been assumed so here. If this assumption is indeed false, wouldn't this proof fail?
Thank you so much!