How is it possible to have a transitive yet not complete relation?

Question

How can we say we cannot compare between pairs (no completeness) and yet we can have transitivity?

Let's assume the relation is a preference relation.

For instance if my set has: $(a, b, c)$ How can I say that a is preferred to b, b is preferred to c and hence a is preferred to c (transitivity) if the relation is not complete and thus I cannot compare a to b and b to c?

Answer 1

EDIT: This question is about microeconomic theory, not set theory. Directly following is an answer related to microeconomics. Further below is a set-theoretic answer, which I'm going to leave there in case someone searches for similar keywords.

Microeconomic Answer:

In economic preference, we say that a relation is complete if and only if for all $A,B$, we have $A \succsim B$ or $B \succsim A$ or both. So consider $A,B,C$ such that $A\succsim B$ and $B \succsim C$. Then the relation is not complete because we do not have an explicit relationship $A \succsim C$. However, we can infer one by transitivity: if $A\succsim B$ and $B \succsim C$, then by transitivity $A \succsim C$, though technically the relation is not `complete'.

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Set-Theoretic Answer:

I don't understand your terms (what do you mean with completeness?). So I am somewhat guessing at what your question actually asks. Please comment with clarifications so I can provide a better answer.

A relation $R$ on a set $X$ is a subset of $X \times X$.

An equivalence relation on $X$ is a relation $R$ on $X$ such that:

• Reflexivity: For all $a \in X, (a,a) \in R$
• Symmetry: For all $a,b \in X$, if $(a,b) \in R$, then $(b,a) \in R$.
• Transitivity: For all $a,b,c\in X$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.

Your notion of completeness seems to be a restatement of trichotomy: in an ordered set, one of the three is true: $a=b$, or $a>b$, or $a<b$.

Notably, we can have transitivity in terms of an equivalence relation on a set, while the set itself is unordered and not transitive. Is this what you mean?

Answer 2

Transitivity says that if $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$, then $\langle a,c\rangle$ must also be in $R$. That’s a conditional statement; when the condition isn’t met, it doesn’t apply and therefore imposes no requirement on $R$.

Suppose that $R$ is a relation on a set $A$. You can think of the elements of $A$ as stepping stones in a river; $\langle a,b\rangle\in R$ means that you can step from $a$ to $b$, while $\langle a,b\rangle\notin R$ means that you cannot. $R$ is transitive if it has the following property:

if you can step from $a$ to $b$ and from $b$ to $c$, then you can step directly from $a$ to $c$.

If you can’t step directly from $a$ to $b$ and from $b$ to $c$, this property tells you nothing about the possibility of stepping directly from $a$ to $c$: maybe you can, and maybe you can’t, and either possibility is consistent with $R$ being transitive.

Suppose, for instance, that the underlying set is $A=\{a,b,c,d,e\}$, and the relation $R$ contains the pairs $\langle a,b\rangle,\langle a,c\rangle,\langle c,d\rangle$, and $\langle a,d\rangle$. You can step from $a$ to $b,c$, or $d$, and you can step from $c$ to $d$. The only two steps that link up are the ones from $a$ to $c$ and from $c$ to $d$: $\langle a,c\rangle\in R$ and $\langle c,d\rangle\in R$.

Is there a shortcut step directly from $a$ to $d$? Yes: $\langle a,d\rangle\in R$. Therefore $R$ is transitive. There’s only one pair of ‘steps’ for which transitivity of $R$ actually requires a shortcut, and the shortcut is present. The fact that $R$ doesn’t even mention $e$ is irrelevant.