MHB Find the error in the following argument

[alexmahone]

Find the error in the following argument by providing a counterexample. “The reflexive property is redundant in the axioms for an equivalence relation. If x ∼ y, then y ∼ x by the symmetric property. Using the transitive property, we can deduce that x ∼ x.”

 

[mathbalarka]

Good problem. This is the "physicist's argument" (no offense topsquark) of the reflexivity principle being redundant, as I was told quite a time ago when I was learning algebra.

The problem with the argument is that you might not have "enough wiggle room" to apply reflexivity. What if the equivalence class of $x$ under $\sim$ is a singleton? How can you say $x \sim y \implies y \sim x$ where there is no such $y$ other than $x$?

Try to think about it a bit. It's quite puzzling for beginners.

 

[Deveno]

The problem with this, is that reflexive means (for a relation $\sim$ on a set $A$):

"For ALL $x \in A$, we have $x \sim x$"

Now certainly if there EXISTS some $y$ with $x \sim y$, we can use symmetry and transitivity to show $x \sim x$. But there is no reason to suppose that we can do this for ANY (thus every) $x \in A$, that is, that such a $y$ even exists.

For example, let $A = \{1,2,3\}$, and let:

$\sim \ = \{(1,1),(1,2),(2,1),(2,2)\}$.

This relation is symmetric, and transitive, but it is not reflexive, since it does not contain $(3,3)$-in fact, NO element of $A$ is related to $3$.