Prove $\mathcal{R}:=\left\{\left(a,b\right)\mid a\le b\right\}$ is not symmetric

Question

Given a homogeneous binary relation $\mathcal{R}$ over a set $A$,and is defied as:

$$\mathcal{R}:=\left\{\left(a,b\right)\mid a\le b\right\}$$

Prove $\mathcal R$ is not symmetric.

---

I don't understand why $\mathcal R$ is not symmetric,indeed it's symmetric if :

$$\forall a,b \in A:a\le b \implies b\le a $$ Or in other words:

$$\forall a,b \in A$$ one of these are true:

$$a < b \implies b< a $$ $$a < b \implies b= a $$ $$a = b \implies b< a $$ $$a = b \implies b= a $$

As it's seen,the last one is true,and that's why I think the relation is symmetric.

Answer 1

$$\forall a,b \in A:a\le b \implies b\le a $$ Or in other words: $$\forall a,b \in A$$ one of these are true: $$a < b \implies b< a $$ $$a < b \implies b= a $$ $$a = b \implies b< a $$ $$a = b \implies b= a $$

You are wrong.

Your second statement is not equivalent to the first. The first statemtent says that for every pair $a,b\in $A, the statement $a\leq b\implies b\leq a$ is true. The statement $a\leq b\implies b\leq a$ is not the same as

$$(a < b \implies b< a) \lor (a < b \implies b= a)\lor (a = b \implies b< a) \lor(a = b \implies b= a) $$

and you can quite clearly show this by taking $a=1$, $b=2$.

---

More generally, your argument is based on assuming that, for any collection of logical statements $A, B, C$, the statement $$(A\lor C)\implies(B\lor C)$$ is equivalent to the statement

$$(A\implies B)\lor(A\implies C)\lor(C\implies B)\lor (C\implies C)$$

but this is not the case.

---

(for clarity, in your argument, $A$ was the statement $a<b$, $B$ was the statement $b<a$, and $C$ was the statement $a=b$)

---

In fact, if you wanted to write out the logical implication (not that I recommend it), then what you would need to consider is that $$(P\lor Q)\implies (R)$$ is equivalent to:

$$\begin{align}(P\lor Q)\implies R&\sim \neg (P\lor Q)\lor R\\&\sim (\neg P \land \neg Q)\lor R \equiv (\neg P\lor R)\land(\neg q \lor R)\\&\sim (P\implies R)\color{red}{\land}(Q\implies R)\end{align}$$

Note the highlighted $\land$ appearing in the statement which is nowhere in your argument.

Answer 2

Symmetry works like this: if $(a,b) \in \mathcal R$, then we always have $(b,a) \in \mathcal R$; is this always true in this case?

For a particular example, consider $(1,2)$, which is in $\mathcal R$ as $1 \le 2$. Then if $\mathcal R$ is symmetric, $(2,1) \in \mathcal R$, but does it meet the condition to be in $\mathcal R$?