Some claims about symmetric relation

Question

On the of symmetric relations, Wikipedia claims:

$\text{A nonempty relation cannot be both symmetric and asymmetric}\tag{I}$

$\text{A relation can be neither symmetric nor asymmetric}\tag{II}$

$\text{A symmetric and transitive relation is always quasireflexive}\tag{II}$

I tried to prove the statements:

$(\text{I})$

Given a binary relation $R$,it's either non-empty or empty,if $R$ is empty then it's vacuously symmetric and asymmetric (indeed it's the only case for which a binary relation is symmetric and asymmetric at the same time),If $R$ is non-empty,then there exist $a,b \in A$ (not necessarily equal) such that if $R$ is symmetric then :

$$(a,b) \in R \implies (b,a)\in R$$ Implies $(b,a) \notin R$ does not hold,iff the relation is not asymmetric.

On the other hand if $R$ is asymmetric then:

$$(a,b) \in R \implies (b,a)\notin R$$ Implies $(b,a) \in R$ does not hold,iff the relation is not symmetric.

From these two it's seen that the two properties does not hold at the same time on a non-empty binary relation.

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$(\text{II})$

An example is the relation defined by :

$$R:=\left\{(a,b) \mid b=a^2:a,b \in \mathbb R\right\}$$

The relation is not symmetric since $(-1,1) \in R$ but $(1,-1) \notin R$,the relation is not asymmetric since $(0,0) \in R$ implies $(0,0) \in R$.

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$(\text{III})$

The last claim is a little ambiguous,one may think the statement claims "a symmetric relation is always quasireflexive" or "a transitive relation is always quasireflexive".

Indeed the claim should be written as:

A relation which is both symmetric and transitive is always quasireflexive.

Given a relation $R$ over $A$ which is both symmetric and transitive,then duo to the symmetric property:

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

Also if $(a,b) \in R $ and $(b,a) \in R$ from the transitive property of $R$ follows $(a,a) \in R$

The same argument does hold for $(b,b)$,therefore we conclude that $R$ is quasireflexive.

It would be appreciated if someone check the proofs.

Answer 1

You can simplify the proof of (I) by proving its contrapositive as follows:

Assume $R$ is both symmetric and antisymmetric. Then

$(a,b) \in R \Rightarrow (b,a) \in R$ since $R$ is symmetric

but also

$(b,a) \in R \Rightarrow (a,b) \not \in R$ since $R$ is antisymmetric

i.e. $(a,b) \in R \Rightarrow (b,a) \in R \Rightarrow (a,b) \not \in R$

So if $R$ is both symmetric and antisymmetric then $R$ must be empty.

Answer 2

I will each statement and then point out how the questioner's proof could be improved.

A nonempty relation cannot be both symmetric and asymmetric.

Proof: let $R$ be a relation. Now suppose we have some $(a, b) \in R$. Then it must be that $(b, a) \in R$, since $R$ is symmetric. And it must be that $(a, b) \notin R$, since $R$ is asymmetric. Contradiction. Then there cannot be any $(a, b) \in R$. Then $R$ is empty. That is, every relation which is both symmetric and asymmetric is empty. Therefore, there is no non-empty, symmetric, and asymmetric relation.

Comment: there is no reason to prove anything about what happens when $R$ is empty in order to prove this theorem.

A relation can be neither symmetric nor asymmetric. Example: $R = \{(1, 1), (0, 1)\}$. Suppose $R$ were symmetric. Now $(0, 1) \in R$. Then $(1, 0) \in R$, since $R$ is symmetric. This is a contradiction. Thus, $R$ is not symmetric. Now suppose $R$ were asymmetric. Now $(1, 1) \in R$. Then $(1, 1) \notin R$. This is a contradiction; then $R$ is not asymmetric.

Comment: The example the questioner provides of a relation which is neither symmetric nor asymmetric is a correct one. However, the questioner attempts to prove that $R$ is not asymmetric by noting that $(0, 0) \in R$ implies $(0, 0) \in R$. This is incorrect reasoning. For consider $R = \emptyset$. Then it can clearly be said that $(0, 0) \in R$ implies $(0, 0) \in R$; nevertheless, $R$ is still asymmetric.

A symmetric and transitive relation is always quasireflexive.

Proof: let $R$ be a symmetric, transitive relation. We write $aRb$ as shorthand for $(a, b) \in R$. We wish to prove that $\forall x ((\exists y (x R y)) \implies x R x)$. Indeed, let an arbitrary $x$ be given, and suppose there is $y$ s.t. $x R y$. Then $y R x$ since $R$ is symmetric. Then since $xRy$ and $yRx$, we have $xRx$ by the transitive property.

Comment: there is no need to also note that $yRy$ to finish the proof, as the original questioner did in the equivalent portion of the posted proof.