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I have a question about equivalence relations:

If I have a relation "$\sim$" which is symmetric and transitive, can I say that if:

$$a,b\in A : a\sim b$$

As "$\sim$" is symmetric:

$$a\sim b \Rightarrow b\sim a$$

And as "$\sim$" is transitive:

$$a\sim b\land b\sim a \Rightarrow a\sim a$$

So, is "$\sim$" reflexive? This prove that if any relation symmetric and transitive is reflexive?

Tryncha
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  • The empty relation (no things relate to each other) is always transitive and symmetric. – k.stm Oct 09 '19 at 07:16

2 Answers2

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It's subtle, but bear in mind the definition of symmetry: if $a \sim b$ then $b \sim a$. (Emphasis on "if" and "then.")

You are not guaranteed that every element is related to some other element. Yes, if for all $a$ there exists some $b$ such that $a \sim b$, then through symmetry and transitivity you can conclude $a \sim a$ and then reflexivity.

But what if there exists some $a$ such that there is no $b$ where $a \sim b$?

A notable example of this is the empty relation on a non-empty set, where no element of the set of concern is related to any other element. (If you view a relation as a subset of the Cartesian product of the set with itself, then the empty relation is the empty set.) You can show that this relation is symmetric and transition through vacuous logic, but is not reflexive. (It is, however, anti-reflexive.)

Some other examples that are less contrived can be found on this post.

PrincessEev
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    If you've seen the proof that there's a $1-1$ correspondence between equivalence relations on a set and partitions of the set, just ask yourself why you need the reflexive axiom for that proof. That will help you reach the correct conclusion on your own. – Robert Shore Oct 08 '19 at 05:02
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On set $A=${$1,2,3$}, the relation $R=${$(1,2),(2,1),(1,1),(2,2)$} is symmetric and transitive but not reflexive.

Nitin Uniyal
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