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Suppose $(A,R)$ be structure where R is a binary relation on $A$. Suppose $A$ has the property that every Non empty subset of $A$ has a least element w.r.t. the relation $R$. Then $R$ is a linear order on $A$.

Proof:

Trichotomy: For any two $x,y \in A$, and $x\neq y$ , consider the two elements subset $\{x,y\}$ of $A$.

Anti-Symmetry: Given $x,y \in A$, $x\neq y$ and $xRy$. Now $yRx$ would imply non existence of least element in $\{x,y\}$.

Transitive: Given $xRy$ and $yRz$.

For $x=y$, it is trivial that $xRz$.

For $x\neq y$:

$x\neq z$, otherwise $yRx$, which contradicts Anti-Symmetry.

Hence by trichotomy, $zRx$ or $xRz$.

For $zRx$ consider the 3-element subset $\{x,y,z\}$, which will not have a least element. Hence $xRz$.

Thus $R$ is a linear order. $\blacksquare$

What is wrong with my reasoning?

This question is inspired by Noah's answer to following question:

https://math.stackexchange.com/questions/2703756/the-difference-between-well-order-and-total-order#:~:text=A%20well%2Dordering%2C%20as%20you,itself%20has%20no%20least%20element).

P.S.: I don't have enough reputation to comment on Noah's answer and ask for clarification.

Arturo Magidin
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  • Trichotomy usually requires that exactly one of $xRy$, $yRx$, and $x=y$ hold. You only proved that at least one does, and that $xRy$ and $yRx$ cannot be the only two that hold. That is not trichotomy. You are going to need to fix something else, because as written your claim would imply that a nonstrict well-order (e.g., $\mathbb{N}$ under $\leq$) is a strict well order, which is false. – Arturo Magidin Mar 03 '23 at 04:48
  • Looking at Noah's answer, I think he said "antisymmetric" when he meant "asymmetric". – Arturo Magidin Mar 03 '23 at 04:55
  • @Arturo. Thank you so much for your help. I got your point $xRx$ and $x=y$ can hold simultaneously. That's not trichotomy. Is it true that with the above premises with additional $R$ being irreflexive, then $R$ is a strict linear ordering. –  Mar 03 '23 at 05:00
  • @ArturoMagidin Yes, you are right. He used asymmetry which different from anti-symmetry. –  Mar 03 '23 at 05:05
  • Your argument that $xRy$ and $yRx$ do not both hold is also wrong. (This is where Noah's assumption of asymmetry comes in). The element $x\in A$ is a least element of $A$ under $R$ if for every $y\in A$, if $x\neq y$ then $xRy$.This definition does not require uniqueness, so the set ${x,y}$ with $x\neq y$ would still have a (in fact, two) least element if both $xRy$ and $yRx$ hold. You either need to assume every nonempty subset has a unique least element and irreflexivity (that will imply asymmetry), or just asymmetry (which implies irreflexivity) as Noah does. – Arturo Magidin Mar 03 '23 at 05:09
  • PS Don't use math mode to produce italic font. Italic font is different from math italic. Use *text* – Arturo Magidin Mar 03 '23 at 05:13
  • Thank you so much for insight that least element need not to be unique. Just for the sum up things, can you give a formal statement: under what minimal conditions (A,R), R will be strict linear order given every Non empty subset of A has least element.(P.S. you guys, like you, Noah, Asaf, bof are my superstars. I have learnt lot about basics of set theory from you). –  Mar 03 '23 at 05:16
  • A counterexample to "every nonempty set has a least element and irreflexive imply linear order" is the two element set ${x,y}$ with $R={(x,y), (y,x)}$, for the reason mentioned above this satisfies both conditions, but is not a (strict) order. – Arturo Magidin Mar 03 '23 at 05:16
  • If (i) every nonempty set has a least element; and (ii) $R$ is asymmetric, then $R$ is a linear (in fact, well-) order. Also if (i) every nonempty set has a unique least element; and (ii) $R$ is irreflexive. Probably other combinations as well. – Arturo Magidin Mar 03 '23 at 05:19
  • @ArturoMagidin I can't thank you enough for being so generous to me and giving your precious time to clarify my misunderstanding. (P.S. I don't understand your italic comment, I am new to latex. So can you please explain it with a simple example , ofcourse if you have time) –  Mar 03 '23 at 05:24
  • Look at the edit I made. You wrote $Proof$ to get what looks like italic font (actually, what is called "math italic"). That produces this: $Proof$. Instead, you should just write *Proof*, which will produce this: Proof. This is done outside of latex/mathjax. Any text between *s is rendered in italic. Use ** to get bold text. **Proof like this** gives Proof like this – Arturo Magidin Mar 03 '23 at 05:26
  • @ArturoMagidin I will keep it in my mind and will not use math italic unnecessarily. Thank you so much. –  Mar 03 '23 at 05:28
  • See edit I was making while you commented. – Arturo Magidin Mar 03 '23 at 05:29
  • @ArturoMagidin I have gone through the edit. Thank you so much for being so kind to me and teaching me some basics of latex. I will keep it in mind in my further posts. –  Mar 03 '23 at 05:34

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