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Suppose $R$ is a strict partial order on $A$. Let $S$ be the reflexive closure of $R$. Show that $S$ is a partial order on A. ("How to Prove it", Chapter 4.5 exercise 4.a)

To prove that $S$ is a partial order on A, I must show the following three properties :

  1. $S$ is reflexive
  2. $S$ is transitive
  3. $S$ is antisymmetric

Since reflexivity is implied in the definition of a reflexive closure, no need to prove anything. To prove property 2 and 3, I assumed that $S = R \cup i_A$, and then I applied proof by cases.

Is the assumption that $S = R \cup i_A$ legimiate in the proof or can it be proven without it?

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Yes, is legitimate and you can quote Theorem 4.5.2 from the text, which says that the reflexive closure is $R\cup i_A$. No need to reprove it again.

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  • $\begingroup$ So $R \cup i_A$ is the reflexive closure for all relations? $\endgroup$ Mar 16, 2015 at 15:50
  • $\begingroup$ Yes, it is, and that's exactly the content of Theorem 4.5.2. $\endgroup$
    – Theo
    Mar 16, 2015 at 15:52
  • $\begingroup$ Yes, it not stated in the Theorem itself, but now that you say it the proof shows it. Thank you! $\endgroup$ Mar 16, 2015 at 15:58
  • $\begingroup$ You are welcome. Indeed, it is not mentioned in the statement, but in the first line of the proof :). $\endgroup$
    – Theo
    Mar 16, 2015 at 16:02

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