Commutative, Transitive but not Reflexive

Wed, 10/28/2009 - 14:25 | by Mgccl

I remember this problem from Apostol's calculus textbook.

Let $\Box$ be a commutative and transitive relation over a non-empty set. Show this proof is flawed:

$a\Box b \implies b\Box a$
$a\Box b \wedge b\Box c \implies a\Box c$
$a\Box b \wedge b\Box a \implies a\Box a$
$\Box$ is also reflexive over the set.

I asked this online like few months ago. One of the people said something about it is related to logic and I should figure it out by looking over the definition of the relations.

I thought of this again, and put it up on Facebook. Thx to Sackel sending me part of this Wikipedia article on equivalence relation.

The empty relation R on a non-empty set X (i.e. aRb is never true) is vacuously symmetric and transitive, but not reflexive.

awesome, now I see how this is not true.
First, here are the definition of the 3 kind of relations.
$\forall_{a \in X} a \Box a$ (reflexive)
$\forall_{a,b \in X} a \Box b \implies b \Box a$ (commutative)
$\forall_{a,b,c \in X} a\Box b \wedge b\Box c \implies a\Box c$ (transitive)

$(a\Box b \wedge b\Box a \implies a\Box a) \Leftrightarrow True$
but that doesn't mean $a \Box a \Leftrightarrow True$ . So the last step of the proof was wrong.

Logic. I see.

Tags:

Math

Comments

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