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Show that the transitive closure of the symmetric closure of the reflexive closure of a relation R is the smallest equivalence relation that contains R.

I can understand the statement intuitively but can't come up with a mathematical proof

Ajay Karthik
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  • Show us this intuitive statement in your head and we will be able to work with you. – Don Larynx Nov 14 '13 at 05:42
  • getting the equivalence relation of any function, i'd start with first finding the reflexive closure, then the symmetric and finally the transitive. This involves the least number of changes, when compared to doing it in any other order – Ajay Karthik Nov 15 '13 at 02:57

1 Answers1

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HINT: Let $S$ be the transitive closure of the symmetric closure of the reflexive closure of $R$. You have to show three things:

  1. $R\subseteq S$.
  2. $S$ is an equivalence relation.
  3. If $E$ is an equivalence relation containing $R$, then $E\supseteq S$.

The first of these is pretty trivial, and the second isn’t very hard: just show that the symmetric closure of a reflexive relation is still reflexive, and that the transitive closure of a symmetric, reflexive relation is still symmetric and reflexive. For (3), show that every ordered pair in $S$ must necessarily belong to any equivalence relation containing $R$.

Brian M. Scott
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