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First, it's my first try in real analysis. I study "Principles of Mathematical Analysis" by Rudin.

In the definition 1.5 of order of a set, I quote:

Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following two properties:

(i) If $x \in S$ and $y \in S$ then one and only one of the statements $x<y, x=y, x>y$ is true.

(ii) If $x,y,z\in S$, if $x<y$ and $y<z$, then $x<z$.

My question is, isn't that property is so obvious with no counter example - as I think - ? why it's so important to state this particular property? why doesn't the second propery being enough for defining order?

Xander Henderson
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Mohamed Mostafa
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  • (ii) doesn't imply (i) in general... what exactly confuses you? – cnikbesku Jan 29 '24 at 17:46
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    There are plenty of relations that fail to satisfy either one of these axioms. There are relations on sets with--for certain points--none of $x<y$, $x=y$, or $x>y$ being true as $x$ and $y$ aren't comparable under the relation – Randall Jan 29 '24 at 17:46
  • @Randall So the confusion comes because I restrict my thinking in numbers? – Mohamed Mostafa Jan 29 '24 at 17:52
  • Yes, that is correct. Look at some of the examples here: https://en.wikipedia.org/wiki/Partially_ordered_set – Randall Jan 29 '24 at 17:53
  • See, for example, https://math.stackexchange.com/q/367583/ . – Xander Henderson Jan 29 '24 at 17:55
  • @Randall So I can replace the first property with less rigorous terms to be "$x$ and $y$ are comparable". Thanks a lot for clarifying :) – Mohamed Mostafa Jan 29 '24 at 18:03
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    Pretty much, yes. If they are comparable they are either equal or one must be greater than the other. The second property says that you cannot have both $x < y$ and $y<x$, for then $x < x$. – Randall Jan 29 '24 at 18:05

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Consider the order $\subsetneq$ on $\mathcal{P}(\mathbb{R})$. It is clearly the case that if $A\subsetneq B$ and $B\subsetneq C$, then $A\subsetneq C$, however, as as example, none of $[0,2]\subsetneq[1,3]$, $[0,2]\supsetneq[1,3]$, $[0,2]=[1,3]$ is true.

Lorago
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