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Kalyan K.
Calculus 3
4 months, 4 weeks ago
Prove that the subsets of a given finite nonempty set X are partially ordered by set inclusion.
in this problem, we have been asked to prove that the subsets of finite non empty set X are partially ordered by set inclusion. So what we need to do is prove that this relation of set inclusion is a fractional order. So, first of all we need to prove that it is reflexive. This relation will be reflexive because let us consider any subsidy of X. And we can say that he is a subset of E. It is a subset of itself and thus is related to enhance. This relation is reflexive. Next we need to show that it is anti symmetric. Now, for that let us consider any two subsets of X, A and B. And let us assume that he is related to B and B is related to A. So that means that is a subset of B, and B is a subset of A. Now if is included in B and bs included an A. Then the only possible conclusion is that the two sets A and B are equal. So, since this condition implies that is equal to be. Hence we can say that the relation is anti symmetric. And the third condition that we need to show is that it is transitive. So for that let us consider three subsets of X, A, B and C. And let us suppose that is a subset of B, and B is a subset of. See. Now, if A is included in B and bs included in C, then we can conclude that is included in C. For example, if we consider a Venn diagram, this is the set E. This is included in the set B, and that's it is included in the set C. So from this diagram we can see that the set E is included in this set. See So this is what we get. So if it is related to B and B is related to see, we get the day is related to. See. Hence it is transitive. Thus, set inclusion is reflexive, it is anti symmetric, and it is also transitive, thus it is a partial order, and thus we approve that the subsets of a given finite non implicit X are partially ordered by set inclusion.
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