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I've been thinking about the following hypothetical problem given in a discrete math textbook (with no explanations), but don't know if I'm going in the right direction.

Given the poset:

⟨{{A}, {B}, {D}, {A, B}, {A, D}, {C, D}, {A, B, D}, {B, C, D}}, ⊆⟩

is it possible to find examples of:

(a) Two elements in the poset that have no lower bound?

  • Here I think that since the elements {A}, {B}, and {D} are sets with a single element, with: {A} and {B}, {B} and {D}, or {A} and {D} as examples of pairs which satisfy this because since they are not necessarily related to one another, they cannot be subsets of one another. However they can be a subset of themselves, as well as the empty set.

(b) Two elements in the poset that have lower bounds, but no greatest lower bound?

  • In the same line of thinking as the previous part, it would have to be both of the elements {A, D} or {C, D}?
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    If the two elements had one common lower bound, is it possible that common lower bound either could or could not be the greatest lower bound? In other words, there is nothing in this problem that is restricting this from happening?2012-11-28

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(a) You’re correct: any two of $\{A\},\{B\}$, and $\{D\}$ are an example of two elements of the poset with no lower bound. The pair $\big\{\{A,B\},\{C,D\}\big\}$ also works, as does the pair $\big\{\{A\},\{B,C,D\}\big\}$.

(b) The elements $\{A,D\}$ and $\{C,D\}$ do have a greatest lower bound, namely, $\{D\}$. However, the pair $\big\{\{A,B,D\},\{B,C,D\}\big\}$ does satisfy the specified conditions: it has $\{B\}$ and $\{D\}$ as lower bounds, and neither of these is a subset of the other, so it has no greatest lower bound.