Is my 'proof' that any strict total order is not total correct?

Question

I am wondering whether a strict total order is total.

As I understand it, a strict total order is (among other things) irreflexive.

But for a relation $R$ to be total, it needs to be true that for any $a$ and $b$; either $aRb$ or $bRa$.

So, if we pick $a=b$, then we must have $aRa$ for any $a$ for it to be total, making it reflexive, rather than irreflexive.

So, it seems that every strict total order is not total.

Is this correct? It seems ... counterintuitive ...

Answer 1

You are quoting the condition of "total" wrong, for strict orders. For the strict order we require the trichotomy law to be:

For all $a,b$ either $a<b$ or $b<a$ or $a=b$.

The last option happens automatically in the non-strict version, as you said, due to reflexitivity. But on the other hand, anti-symmetry is easier to formulate for strict orders (and in fact follows from irreflexivity and transitivity).