Just a question I was having some trouble with, any thoughts would be appreciated.
- Suppose S is a (finite) set containing at least two elements for A,B that are elements of P(S), we define A is a poset on B to mean |A| =< |B|. Is the relation a partial order on P(S) and explain.
Thanks!
to determine if we have a partial ordering, we must verify that the set has three properties: reflexivity, anti-symmetry and transitivity. that is, we must show that:Originally Posted by kevi555
1.for all
2. For all, If
and
then
3. For all, If
then
and it seems that you have defined
by the way, what does "poset" mean?
i don't get what you are saying. i'm not familiar with the symbol you are talking about. you mean the less than or equal sign with a wavy line under it? in any case, i think that's just giving us a criterion by which to compare any two elements in the set in terms of therelation.
i stick by my answer. check if the properties i've mentioned hold.
It appears that you are confused about the terminology of ‘posets’.
I think what the problem surely must be is the following
Suppose S is a (finite) set containing at least two elements, define a relation on P(S) by this ARB if and only if |A|<|B|. Now the question is if (P(S), R) is a partial ordering or a poset?
Or, is R a partial order on P(S)?
Now clearly ordinary comparison of counting numbers is both reflexive and transitive.
Your task is to decide if it is antisymmetric.
HINT: If I were you, I would look for a simple counterexample.
Post Script to Jhevon.
|A| is the cardinality of the set A.
‘poset’ is a standard term, it means partially ordered set.
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