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We must show that there exists some kind of $\alpha$ relation
$\alpha ⊆ X \times X$ which has these conditions :

if this relation is I and II type.
I) symmetrical:
if $∀x,x' ∈ X : (x, x') ∈ \alpha ⇒ (x', x) ∈ \alpha$
II) transitive:
if $∀x, x', x'' ∈ \alpha : (x, x') ∈ \alpha (x',x'') ∈ \alpha ⇒ (x, x'') ∈ \alpha$

then there must be result of non-reflexivity for this $\alpha$ relation
III) nonreflexive:
if $∀x ∈ X : (x, x) $ $\notin \alpha$

Is there any kind of relation like this?

Martin Sleziak
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IremadzeArchil19910311
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1 Answers1

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If $X=\{x\}$ or $X=\emptyset$, then it's easy to show that $\alpha=\emptyset$.

Suppose $X$ has more than one element. Take $x\ne y$ - elements of $X$. Suppose also that $(x,y)\in\alpha$, then $(y,x)\in\alpha$, hence $(x,x)\in\alpha$, which leads to contradiction.

Thus, $\alpha=\emptyset$.

TZakrevskiy
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