# Proof of every asymmetric relation is irreflexive

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!

But it is not assumed to be reflexive. Remember that reflexive means for all $a\in A$ it holds that $a<a$. Here we only assume that for some $a\in A$ it holds that $a<a$, which is exactly what it means not to be irreflexive.
Definition. A relation that is irreflexive, is a binary relation on a set where no element is related to itself; in other words: $$\forall a\in A; \lnot (a<a)$$
Let < not irreflexive. So $$\lnot (\forall a\in A; \lnot (a<a))$$ which means $$\exists a\in A,a < a.$$ And this is what you want: "the relation is not irreflexive, there is an $a∈A$ with $a < a$"