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Drawing hasse diagram for the poset ({{1}, {2}, {4},{1, 2}, {1, 4}, {2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}}, ⊆).

a) The maximal elements are the ones without any elements lying above them in the Hasse diagram, namely {1,2}, {1,3,4}, and {2,3,4}.

b) The minimal elements are the ones without any elements lying below them in the Hasse diagram, namely {1} , {2} ,and {4} .

c) There is no greatest element, since there is more than one maximal element, none of which is greater than the others.

d) There is no least element, since there is more than one minimal element, none of which is less than the others.

e) The upper bounds are the sets containing both {2} and {4} as subsets, i.e., the sets containing both 2 and 4 as elements. Pictorially, these are the elements lying above both {2} and {4} (in the sense of there being a path in the diagram), namely {2,4} and {2,3,4}.

f) The least upper bound is an upper bound that is less than every other upper bound. We found the upper bounds in part (e), and since {2,4} is less than (i.e., a subset of) {2,3,4}, we conclude that {2,4} is the least upper bound.

g) To be a lower bound of both {l, 3, 4} and {2, 3, 4}, a set must be a subset of each, and so must be a subset of their intersection, {3, 4}. There are only two such subsets in our poset, namely {3, 4} and { 4}. In the diagram, these are the points which lie below (in the path sense) both {1,3,4} and {2,3,4}.

h) The greatest lower bound is a lower bound that is greater than every other lower bound. We found the lower bounds in part (g), and since {3,4} is greater than (i.e., a superset of) {4}, we conclude that {3,4} is the greatest lower bound.

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