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Given a partial ordering $R$ over a set $S$ is it true that for every $A\subseteq S$ that $R$ is also a partial ordering over $A$? I think so but I'm not sure.

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More precisely, $R\cap(A\times A)$ is a partial order on $A$. –  Brian M. Scott Feb 9 '12 at 8:51
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up vote 4 down vote accepted

Yes, it is. The three defining properties of a partial order, reflexivity, antisymmetry and transitivity, contain only universal quantifiers and no existential quantifiers, and therefore can't be broken my removing elements from the set.

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While it seems pretty obvious, is it something I would need to prove or is it so obvious that I could rely on it without proving it? –  Robert S. Barnes Feb 9 '12 at 10:56
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In an elementary first course, prove it. In a research paper, assume your reader knows it. –  GEdgar Feb 9 '12 at 13:31
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