Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
More precisely, $R\cap(A\times A)$ is a partial order on $A$. – Brian M. Scott Feb 9 '12 at 8:51
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.

share|cite|improve this answer
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
In an elementary first course, prove it. In a research paper, assume your reader knows it. – GEdgar Feb 9 '12 at 13:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.