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Let $L$ be a set with a relation $R$ which is transitive, anti-symmetric and reflexive and for any two elements $a, b \in  L$, let the least upper bound $lub (a, b)$ and the greatest lower bound $glb (a, b)$ exist. Which of the following is/are true?

  1. $L$ is a poset

  2. $L$ is a Boolean algebra

  3. $L$ is a lattice

  4. None of the above

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16 votes

Which of the following is/are true? This is question with Multiple answers.

As our Relation $R$ on Set $L$ is Reflexive , anti symmetric & Transitive it is poset.

Since $\text{LUB}$ & $\text{GLB}$ exists for any two elements it is lattice.

Answer: $A$ & $C$.

$B$ is not guaranteed to be true.

Ref: http://uosis.mif.vu.lt/~valdas/PhD/Kursinis2/Sasao99/Chapter2.pdf

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it is given that set with a relation Relation R which is transitive, antisymmetric and reflexive so it is already a poset .Additionally, it is guaranteeing about the existence  of LUB and GLB too.so now it is  : Poset + Having lub and Glb both , That is Lattice . so option A and C both r true .

For being option B true it needs some extra information that is :

A lattice can be Boolean Algebra only  when It is distributed lattice , complimented lattice and bounded lattice along with it will have to satisfy the properties of Boolean algebra . and we dont have any such info about this lattice .

counter of B is :

  it is poset, lattice but not B.A

1 votes
1 votes
As it is Reflexive , anti-symmetric , transitive , it is poset.

Since lub and gub exist for any pair of elements, the poset becomes a lattice. So, C is the answer.
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0 votes
The answer is a and c. because in the question it is given that LUB and GLB exists and by definition LUB and GLB are always unique. If it is not unique that we can say that LUB and GLB does not exist. It is not required to mention explicitly the word unique.
Answer:

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