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 ...


1 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).

  • $\begingroup$ Sorry, I am not quite following you. Is it "No, a strict order is not total", or is it "No, your proof that a strict total order is not total is not correct"? I should probably update my title to avoid this confusion :) $\endgroup$
    – Bram28
    Apr 16, 2018 at 16:52
  • $\begingroup$ Is it better now? $\endgroup$
    – Asaf Karagila
    Apr 16, 2018 at 16:53
  • $\begingroup$ Yes, thanks! OK, so we are using a different notion of 'totality' when talking about total orders as opposed to when we are talking about strict total orders? Why not use something like: For any $a \not = b$: either $aRb$ or $bRa$ as the condition for both kinds of total orders? Also: since strict total orders don't have the condition of 'totality' as defined for total orders, can't I still argue that strict total orders are not total orders? By the way, I am using the Wikipedia page on total orders here; maybe that's my cardinal sin :) $\endgroup$
    – Bram28
    Apr 16, 2018 at 17:30
  • $\begingroup$ Yes, many properties of partial orders have strict and non-strict variants, which sometimes become easier in one way and sometimes in another. It is customary to decide in advance, what is the "correct context", and follow those definitions for the most part. For introductory material, this is mostly the reflexive version, for more advanced uses, it is often the strict version that works better. $\endgroup$
    – Asaf Karagila
    Apr 16, 2018 at 17:49
  • $\begingroup$ Interesting, thanks! But still: why not use trichotomy for both? $\endgroup$
    – Bram28
    Apr 16, 2018 at 19:40

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