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

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.

share|cite|improve this question
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
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 $\subseteq$.

share|cite|improve this answer
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
@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

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.