# Subset of a partially ordered set: Least upper bound and greatest lower bound.

Let $(A, \leq)$ be a poset and $B \subseteq A$. I need to show

i) B may have at most one least upper bound in A

ii) B may have at most one greatest lower bound in A.

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This is really a matter of unpacking and applying the definitions of what it means

• for $A$ to be a poset (poset: "partially ordered set")

• (See also the section of that entry subtitled Extrema)
• for $B$ to be a subset of $A$,
• for a set to have a least upper bound (lub) and
• for a set to have greatest lower bound (glb).

Recall the definition of a least upper bound as:

Let S be a poset:

1. $\alpha$ is an upper bound for S if $x \leq \alpha \space \forall x \in S$, (upper and lower bounds are NOT unique) and
2. $\beta$ is the least upper bound for S if $\beta$ is an upper bound, and $\beta \leq \alpha$ whenever $\alpha$ is an upper bound for S.

A symmetrical definition defines the greatest lower bound.

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Sorry i m a newbie with this. I get the least upper bound but cant it it more than 1 least upper bound? –  Ahmed Apr 14 '13 at 17:01
The least upperbound and the greatest lower bound are unique. But there can be infinitely many "upper" or "lower" bounds. We're interested here in the least/greatest (respectively) bounds. –  amWhy Apr 14 '13 at 17:08
Also note that you are asked to show that $B$ has at most one least upper bound in A, at most one greatest lower bound in A. It may be the case that no such lub or glb for $B$ exists in $A$, but if either does exist in A, then it is the unique lub/glb of $B$ in A, respectively. Make sure you're clear about such bounds for $A$, given it's a poset. Then figure out what this means with respect to lub/glb bounds for $B\subset A$. –  amWhy Apr 14 '13 at 18:10
How are you coming on this question, Ahmed? –  amWhy Apr 14 '13 at 19:12
still confused i m reading book right now. But still trying. –  Ahmed Apr 14 '13 at 19:14
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