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The relation is $R = \{(x, y) \in\Bbb R^2 \mid x < y \}$

How can I prove antysimetric property for the relation?

Antisymmetric property states that IF $x R y$ and $y R x$, THEN $x = y$. The thing is, in this relation, the first part of the property is never true, so that's how I know that the relation IS antisymmetric. I just don't know how to prove it.

zipirovich
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tobi
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    Edit the question to show us what you tried, or at least what you know. What assertion would have to be true to prove antisymmetry? – Ethan Bolker Jul 09 '18 at 18:07
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    Yes, that's the thing. Antisymmetric property states that IF x R y and y R x, THEN x = y. The thing is, in this relation, the first part of the property is never true, so that's how I know that the relation IS antisymmetric. I just don't know how to prove it. – tobi Jul 09 '18 at 18:10
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    $P \rightarrow Q$ is true provided $P$ is false. And so the if-then statement you have is true: it cannot be that $x R y$ and $y R x$, for this says $x < y$ and $y < x$, which yields $x < y < x$: a clear contradiction as $x \not < x$. – Benjamin Dickman Jul 09 '18 at 18:16
  • For a proof, how do you define $<$? – Hagen von Eitzen Jul 09 '18 at 18:49

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[Migrating a comment to serve as an answer.]

A reminder about logical implications of the form "If $P$, then $Q$" and denoted $P \rightarrow Q$:

Such an if-then statement is true if either both $P$ and $Q$ are true, or if $P$ is false.

You are in the latter case; as you write:

Antisymmetric property states that IF x R y and y R x, THEN x = y. The thing is, in this relation, the first part of the property is never true, so that's how I know that the relation IS antisymmetric.

Indeed, it cannot be the case that $xRy$ and $yRx$, for this would mean we had $x, y \in \mathbb{R}$ for which both $x<y$ and $y<x$; but, those inequalities yield $x < y < x$, which is a contradiction: $x \not < x$.

Since the antecedent (i.e., $P$) is false, the implication (i.e., $P \rightarrow Q$) is true. As one more piece of vocabulary: In such a scenario, one sometimes says that the statement is vacuously true, i.e., when an implication is true because its antecedent is false.

Benjamin Dickman
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