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Let $Y$ be a set and suppose$P \subseteq \mathcal{P}(Y)$. Show that $P$ is partially ordered set by $\subseteq$.

So I start off with $\mathcal{P}(Y)=\{\emptyset\subseteq,......\subseteq P....\subseteq..\subseteq Y\}$

How should I use now the reflexivity, antisymmetry and transitivity? I don't think I understand the question. The inclusion is a partial ordering relation on the power set but how to show that one of it's members is partially ordered?

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You are only supposed to verify that this is a partial order. All verifications will be easy. It has nothing to do with a member being ordered. – André Nicolas Apr 28 '13 at 18:33
Do you know, for a start, that $\mathcal{P}(Y)$ is partially ordered by inclusion? Then the restriction of a partial order is still a partial order. – 1015 Apr 28 '13 at 18:33
You're showing that the power set of Y is a poset under inclusion. Take two elements from teh power set and prove the three... – Eleven-Eleven Apr 28 '13 at 18:37
up vote 2 down vote accepted


$$===\;\;\;\;\;\forall\,A\in P\;,\;\;A\subset A$$

$$===\;\;\;\;\;\forall\,A,B\in P\;,\;\;A\subset B\,\,\wedge\,\,B\subset A\implies A=B$$

$$===\;\;\;\;\;\forall\,A,B,C\in P\;,\;\;A\subset B\;\;\wedge\;\;B\subset C\implies A\subset C$$

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@user74823, I think you're addressing Julien in the wrong part of the thread... – DonAntonio Apr 28 '13 at 18:46
Ah sorry...I am beginner on this side.. So you propose to prove it in exactly the same way as we prove that the power set is ordered by inclusion. The only difference is that we choose subsets of our set P and follow the requirements right? – H.E Apr 28 '13 at 18:54
Of course, @HeidiE . What's the difference between $\,P\,$ and $\,\mathcal P(Y)\,$ ? – DonAntonio Apr 28 '13 at 18:56
well $P\subset \mathcal{P}(Y)$ so subsets of $P$ will be also subsets of $\mathcal{P}(Y)$ is that what you are referring to? – H.E Apr 28 '13 at 18:59
What I meant is as long as we have a set of subsets the $\,\subset\,$ relation with partial order that set. – DonAntonio Apr 28 '13 at 19:02

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