A relation is asymmetric if and only if it's both antisymmetric and irreflexive

Question

Prove

A relation is asymmetric if and only if it's both antisymmetric and irreflexive

Given a relation $\mathcal R$ on a set $A$ (a homogeneous binary relation),then:

$\mathcal R$ is antisymmetric if :

$$\forall a,b \in A:(a\mathcal R b \wedge b \mathcal R a)\implies a=b$$

$\mathcal R$ is irreflexive if : $$\forall a \in A:a\not \mathcal R a $$

$\mathcal R $ is asymmetric if : $$\forall a,b \in A:a \mathcal R b \implies b\not \mathcal R a$$

$\Longleftarrow$

This direction can be proved using contradiction argument,

Assume $\mathcal R$ is both antisymmetric and irreflexive,but it's not asymmetric ,then:

$$\exists a,b \in \mathcal R :a\mathcal R b \implies b \mathcal R a$$

Therefore: $$\exists a,b \in \mathcal R :(a\mathcal R b \wedge b\mathcal R a)$$

On the other hand by the asntisymmetric property of $R$:

$$\forall a,b \in A:(a\mathcal R b \wedge b\mathcal R a)\implies a=b$$

Which means :

$$a\mathcal R b \implies a\mathcal R a$$

Which is not true duo to the irreflexive property of $\mathcal R $,implies if $\mathcal R $ is both antisymmetric and irreflexive,then it's asymmetric.

$\Longrightarrow$

If $\mathcal R$ is asymmetric,then:

$$\forall a,b \in A:a\mathcal R b \implies b\not \mathcal R a$$

One may say take $a=b$ and it follows that $\mathcal R$ is irreflexive,but I don't know how $$\forall a \in A:a\mathcal R a\implies a\not \mathcal R a$$

Does make sense.

Also I don't know how to show the asymmetric property of $\mathcal R$ implies the antisymmetric property.

Answer 1

Your proof is correct, but it's a bit hard to read, you should clarify in a sentence that $a, b$ are taken so that they disprove asymmetry, i.e. $aRb\land bRa$.

For the other direction, antisymmetry follows vacuously from asymmetry, as the condition $aRb\land bRa$ is always false.

For irreflexibility, we indeed use asymmetry with $a=b$, yielding $$\forall a:\ aRa\implies a\not Ra$$ which implies $\forall a:\, a\not Ra$, since if we had $xRx$ for an $x$, it would imply $x\not Rx$, a contradiction.

Answer 2

Let's walk through it.

Suppose $R$ is asymmetric.

Suppose we have $aRb$ and $bRa$. Because $R$ is asymmetric and $aRb$, we have $\neg(bRa)$. This is a contradiction; it contradicts the fact that $bRa$. Since everything follows from a contradiction, we have $a = b$. Then $R$ is antisymmetric.

Suppose we have $aRa$. Then we have $\neg (aRa)$. Contradiction. Then $\neg(aRa)$. Then $R$ is irreflexive.

Thus, asymmetric implies antisymmetric and irreflexive.

Now, on the other hand, suppose $R$ is antisymmetric and irreflexive. Suppose that $aRb$. Now suppose that $bRa$. Then $a = b$, by antisymmetry. Then $aRa$. Contradiction, by irreflexivity. Then $\neg(aRb)$. Thus, $R$ is asymmetric.

Thus, antisymmetric and irreflexive implies asymmetric.

I hope this clarifies things.