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Is the relation $\mathcal{R}=\{(1,2),(2,3),(1,3)\}$ on $X=\{1,2,3\}$ complete?

By looking at the completeness definition in preference: Definition 1.1(c), this is same as the connected relation in the maths definition wiki-connected relation. If I am right, for any distinct pair of elements in $X$, they are comparable, otherwise they are the same elements [They are the same might just leave the chance that $(1,1)\notin\mathcal{R}$?].

In my example, I think this satisfied complete preference(connected relation), even though in the context of daily language that any good should be weakly prefer to itself. Please correct me if this is not connected relation, from definitional or logical disprove.

This question pondered by a question asking to show that complete binary relation on$X$ is also reflexive.

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In the wiki you will see it says: "As described in the terminology section below, the terminology for these properties is not uniform."

The definition of strongly connected in the wiki is the same as the definition of completeness at your first link. Both require that:

$$\text{for all $x,y\in X$ either $(x,y)\in\mathcal{R}$ or $(y,x)\in\mathcal{R}$ }$$

Note that the above must hold when $x=y$ as well as when $x\neq y$.

The definition of connected / complete / total in the wiki is a weaker condition:

$$\text{for all $x,y\in X$ with $x\neq y$ either $(x,y)\in\mathcal{R}$ or $(y,x)\in\mathcal{R}$ }$$


For your relation, since, for example, $(1,1)\notin\mathcal{R}$, the relation is not complete (according to the definition in your first link) and not strongly conected (according to the definition in the wiki). However, your relation $\mathcal{R}$ is connected (according to the definition in the wiki).

The relation $\mathcal{S}=\{(1,1),(2,2),(3,3),(1,3),(2,1),(3,2)\}$ is strongly connected (according to the wiki definition) and therefore connected (according to the wiki definition) and complete (according to the definition at your first link).


If a relation is complete by the definition at your first link, then it is reflexive.

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  • $\begingroup$ Thank you for your reply. My first link $\forall x,y\in X$, either $(x,y)\in\mathcal{R}$ or $(y,x)\in\mathcal{R}$ or (both) is same as the connected relation (not strongly connected) in the wiki. I believe confusion comes in how to interpret $x=y$ in the definition. When you read this definiton, why would perceiving $x=y$ means this requirements need to be hold for $x=y$, whereas not the case that as long as $x=y$ we may ignore that part. Just talk about first link definiton, and wiki connected relation definition. not strongly connected $\endgroup$
    – LJNG
    Aug 16, 2023 at 14:19
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    $\begingroup$ No, the definition of connected in the wiki specifies $x\neq y$, while the definition of complete in your first link does not (so the definitions are not the same). The definition of complete in your first link corresponds to the definition of strongly connected in the wiki. I am not sure what you mean by "how to interpret $x=y$". It is not a matter of interpretation; the two elements $x$ and $y$ are the same or they are not. $\endgroup$
    – smcc
    Aug 16, 2023 at 14:38
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    $\begingroup$ No, that definition of connected is not the same as the definition of complete in your first link. The definition of connected allows for it not to be the case that $x\mathcal{R} x$, whereas the definition of complete requires that $x\succsim x$. $\endgroup$
    – smcc
    Aug 16, 2023 at 15:17
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    $\begingroup$ The formal definitions of completeness (at the first link) and strong connectedness both require that $x$ is related to itself (for all $x\in X$). That is the crucial difference to connectedness. Strong connectedness implies connectedness, but not conversely. It is common sense that every consumption bundle is weakly preferred to itself. That is why strong connectedness is assumed rather than just connectedness. Also, indifference ($x\sim y$) can be defined as $x\succsim y$ and $y\succsim x$, so if it were not the case that $x\succsim x$, then it would not be the case that $x\sim x$. $\endgroup$
    – smcc
    Aug 16, 2023 at 15:31
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    $\begingroup$ In mathematics, "A or B" includes the possibility that "A and B". There is no need to write "A or B or (A and B)". $\endgroup$
    – smcc
    Aug 16, 2023 at 15:50

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