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The problem I am working on is, Find two incomparable elements in these posets.

a) $(P(\{0,1,2\}),⊆)$

b) $(\{1,2,4,6,8\},|)$

For a, I said that $R \subseteq p(\{0,1,2,3\}) \times p(\{0,1,2,3\})$, where $A$ and $B$ are sets, that are elements of the powerset. Then, $R=\{(A,B)|A \subseteq B\}$. An example of two incomparable elements would be $\{0\}$ and $\{1\}$, because they are not subsets of each other. So, the ordered pairs $(\{0\},\{1\})$ and $(\{1\},\{0\})$ are two ordered-pairs that contain elements incomparable to each other. (Is that proper to say that?)

Would this be an acceptable answer? I don't like my textbook's solution: they never use any notation; there answer is completely descriptive, which is nice, but I would like if they supplemented the description with notation.

I don't need help with part b, because if I answered part a correctly, then I will have answered part b correctly.

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Brian Scott has already answered your question, so I just want to add a side remark about notation. Notation is just a concise way of expressing an idea. Your goal should always be for a completely descriptive knowledge of a given problem; of course, as a matter of practicality, you may well prefer to use a concise notation to write your answer. Now there is some value to notation in that well-chosen notation can help clarify ideas in our heads, but it is the clarification of the ideas and not the notation itself that has true value. –  Michael Joyce Nov 15 '12 at 16:10

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up vote 4 down vote accepted

$\{0\}$ and $\{1\}$ are indeed incomparable elements of $\wp(\{0,1,2\})$ with respect to the partial order $\subseteq$, and for the reason that you gave: $\{0\}\nsubseteq\{1\}$, and $\{1\}\nsubseteq\{0\}$. There’s no reason to look at the ordered pairs, though it’s true that neither $\langle\{0\},\{1\}\rangle$ nor $\langle\{1\},\{0\}\rangle$ belongs to the order $\subseteq$.

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Oh, so the relation only gives the type of condition we are looking at when we take the cross product of the powerset? We don't actually look at elements in the relation? –  Mack Nov 15 '12 at 16:02
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@EMACK: I’m not sure what you mean. $\{0\}\nsubseteq\{1\}$ and $\langle\{0\},\{1\}\rangle\notin\subseteq$ are two ways of saying the same thing, and the first one is much more readable! –  Brian M. Scott Nov 15 '12 at 16:06

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