Proof of $\;\text{Asymmetric}(\sqsubset)\rightarrow \text{Antireflexive}(\sqsubset)$

Question

The relation $\;\sqsubset\;\subseteq S\times S$ is asymmetric if

$$\forall a,b\in S:(a,b)\in\sqsubset\rightarrow (b,a)\notin\sqsubset$$

and it is antireflexive if

$$\forall a\in S:(a,a)\notin\;\sqsubset$$

I want to prove that

$$\text{Asymmetric}(\sqsubset)\rightarrow \text{Antireflexive}(\sqsubset)$$

Now... it seem me obvious that is right, eve using some examples in my mind

but when I try to write the formal proof in this way I'm in confusion at the conclusion.

$(a,b)\in\;\sqsubset\rightarrow (b,a)\notin\;\sqsubset$

if $b=a$ I get

$(a,a)\in\;\sqsubset\rightarrow (a,a)\notin\;\sqsubset$

... this is a contraddiction..but I don't understand how it is the proof...

someone can explain me in easy words why this is a proof? I only see that the asymmetry lead to contraddiction when $b=a$ ... there is something I'm missing..

Answer 1

$\text{We take as given that the relation is asymmetric.} \tag{Premise}$

We are assuming, for the sake of contradiction, that $\sqsubset$ is NOT antireflexive; that is we assume that there is an $a \in S$:

$$(a,a)\in\,\sqsubset\tag{1}.$$

Since the relation $\;\sqsubset\;$ is asymmetric, then by definition of asymmetry, $$(a,a)\in\,\sqsubset\rightarrow (a,a)\notin\,\sqsubset\tag{2}.$$

By modus ponens, given $(1),\;\text{and}\; (2)$, we are forced to conclude, therefore $$(a, a) \notin \,\sqsubset\tag{3}$$

$(3)$ contradicts $(1)$, so it cannot be the case that in any asymmetric relation, there exists an $a \in S$ such that $(a, a) \in \sqsubset$. That is, our assumption $(1)$ is false, because it leads to a contradiction. Hence a asymmetric relation must necessarily be antireflexive.

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Note, what leads to the contradiction is the assumption that $\sqsubset$ contains at least one element that is related to itself, while also being asymmetric. We've shown an asymmetric relation cannot NOT be antireflexive, i.e., an asymmetric relation is necessarily antireflexive.

Answer 2

Imagine to display the elements in the cartesian plane, then being asymmetric means that if a point is in the relation, then its symmetric with respect to the bisectrix of first quadrant is not. Taking the point on the bisectrix gives antireflexivity

Answer 3

Let $\sqsubset$ be an asymmetric relation on $S$. Let $a\in S$ be arbitrary. Two cases are possible: $a\sqsubset a$ or $a\not\sqsubset a$.

• If $a\sqsubset a$, then by asymmetry $a\not\sqsubset a$
• If $a\not \sqsubset a$, then we have mmediately that $a\not\sqsubset a$

In both cases we find $a\not\sqsubset a$. Hence $\sqsubset $ is antireflexive.