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.
Best Answer
$\{0\}$ 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 $\subseteq$.