Similar in spirit to a question here, why do we say as part of the axioms for real numbers that: $$\text{"The order relation $<$ has the least upper bound property."}$$ when (presumably) we mean: $$\text{"The set of real numbers $\mathbb{R}$ equipped with the usual order $<$ has the least upper bound property."}$$

To me, it seems wrong or maybe slightly misleading to say the order relation itself has the l.u.b. property since the order relation is a subset of $\mathbb{R}^2$ in this case, or a Cartesian product generally.

Is this reasoning correct, and is the original statement essentially an abuse of terminology?

  • $\begingroup$ What's the potential misinterpretation you're worried about? $\endgroup$
    – Karl
    Commented Aug 6, 2023 at 19:53
  • $\begingroup$ @Karl well an order relation on a set $A$ is itself a subset of $A\times A$, so if we say that a subset of $A\times A$ has the least upper bound property...with respect to what order? It wouldn't necessarily be the order relation on $A$ we started with. $\endgroup$
    – user689775
    Commented Aug 6, 2023 at 19:56
  • $\begingroup$ The original statement is not an abuse of terminology. The field of an order can always be recovered from the relation so the least upper bound property of an order on $\Bbb{R}$ is a property of the order as a subset of $\Bbb{R}^2$. From your comment, you seem to be confusing the subset $<$ of $\Bbb{R}^2$ with $\Bbb{R}^2$ itself. $\endgroup$
    – Rob Arthan
    Commented Aug 6, 2023 at 19:58
  • $\begingroup$ We know that the order relation on $\mathbb{R}$ is a total order, meaning that any two distinct elements in $\mathbb{R}$ are related, in particular this means that the entirety of $\mathbb{R}$ is contained in the projection of $<$ onto either coordinate, so any property of $(\mathbb{R},<)$ can be inferred from $<$ alone, given that $<$ is a total order. Similar to how could deduce any property of a set together with an isomorphism on it, from the isomorphism alone, provided you know that it is an isomorphism. Does that help? $\endgroup$
    – Carlyle
    Commented Aug 6, 2023 at 20:00
  • 1
    $\begingroup$ @upanddownintegrate He takes a shortcut with notation because he supposes a set $R$ on the previous page. All of the properties, including property $7$, are properties of $R$ so, although I agree that he is abusing the language a little, he is saying exactly what you suggested he should say: the set $R$ with order relation $\lt$ has the least upper bound property. $\endgroup$
    – John Douma
    Commented Aug 6, 2023 at 20:39

2 Answers 2


In your context, the order relation of course is a binary relation, so indeed is a subset of $\mathbb{R}^2$. Then, $(\mathbb{R},<)$ has the least upper bound property. It's just a short form to say what you stated in the second centered line.

If we were studying abstract algebra or topology, for instance, where we use less usual sets, and different kind of orders, we clarify it.


The second statement is implicit in the first.

The first statement is neither wrong nor misleading. Saying that a relation has a property implicitly refers to the set on which the relation is defined.

I would not call it an abuse of terminology.

If you are more comfortable making the set explicit, do that when you write, or fill it in when you read.

  • $\begingroup$ I don't think that's quite right (or at least it's context-dependent). In a set-theoretic context (rather than in the category of sets and relations), you can't determine the intended field from the relation in general, but if you say the relation is an order, you are implying that the intended field can be recovered as the union of the domain and range of the relation and the l.u.b. property is a property of the order alone. $\endgroup$
    – Rob Arthan
    Commented Aug 6, 2023 at 20:01
  • $\begingroup$ The order $\lt = \{(a,b)\in\mathbb R^2: a\lt b\}$ is an open half plane. How is that the same thing as $(\mathbb R,\lt)$? I believe @upanddownintegrate is correct. $\endgroup$
    – John Douma
    Commented Aug 6, 2023 at 20:12
  • 1
    $\begingroup$ @JohnDouma: no-one is saying that ${<} \subseteq \Bbb{R}^2$ is the same thing as $(\Bbb{R}, {<})$, but you can recover the latter from the former given the assertion that $<$ is an order, so there is nothing wrong with the original statement. $\endgroup$
    – Rob Arthan
    Commented Aug 6, 2023 at 20:23

You must log in to answer this question.