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Let ~ be a symmetric and transitive relation on a set A. What is wrong with the folloing "proof" that $\sim$ is reflexive?

Proof: $a\sim b$ implies $b\sim a$ by symmetry; then $a\sim b$ and $b\sim a$ imply that $a\sim a$ by transitivity, thus $a\sim a$.)

Thomas Andrews
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Leon K
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1 Answers1

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We must have $\forall a \in A$, $a \sim a$.

You just proved that if $\exists b$ s.t. $a \sim b$, then $a \sim a$. Why is there such $b$?

For example, the empty relation satisfies symmetry and transitiveness, but the lack of existence of something to "apply" them yields non-reflexivity.

Aloizio Macedo
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  • In the book I am studying algebra, happens that following the aforementioned question, comes this:

    Can you suggest an alternative of property 1 [reflexivity] which will insure us that properties 2 [symmetry] and 3 [transitiveness] do imply property 1?

    The first question I answered saying that we can consider an $x$ in $A$ such that there is no $y$ that satisfies $x$ ~ $y$, hence there is no guarantee that reflexivity follows necessarily from the other two properties. But I can't imagine such an alternative as asked in this second item.

     – izzorts Feb 20 '17 at 22:16
  • @izzorts Sorry for the late response. I think the alternative that your question wants is requiring that every $a \in A$ is related to someone (from which reflexivity follows, as the "flawed" argument now applies). Another alternative is requiring that there exists a function $f \subset R$, where $R$ is the relation (both requirements are equivalent assuming choice). – Aloizio Macedo Mar 08 '17 at 03:57