The least upper bound property says that, "Every nonempty subset of $A$ that $is$ bounded above has a least upper bound." The great lower bound property is defined similarly, and it's not difficult to show that one property holds if and only if the other holds. The key lies in the assumption that we only deal with subsets that are bounded below to show they must have an inf.

On the other hand, an old homework (question 5) of mine says to prove that if every subset of a poset has an inf then every subset must also have a sup; I just don't believe this because one must first prove that such subsets are even bounded above (and this was not stated in the problem).

There's also this post on Stack Exchange which has no answer, though the OP believes to have found an answer, it doesn't look good.

Is it fair to say that the homework question is just wrong and what was intended to be said is something like, "Prove the equivalence of the lub and glb properties."?

  • $\begingroup$ Yes, I agree. My point is this: I think the intent was to ask us to show the equivalence of lub and glb properties. $\endgroup$ – Squirtle May 29 '15 at 19:36
  • $\begingroup$ Did you read Ittay's comment on the linked post? $\endgroup$ – anon May 29 '15 at 19:45
  • $\begingroup$ Yes, but I was a little confused by it. I'm assuming that if the set of least upper bounds for a subset $S$ is empty then inf($\emptyset)=\sup(A)$ and this is an upper bound for $S$, contradicting that the set of upper bounds is empty. It just doesn't quite feel right, I realize $\emptyset\subset A$ and so $\inf(\emptyset)$ exists by our assumption, but I suppose I don't see why $\sup(A)$ exists, if it exists then I believe they are equal. $\endgroup$ – Squirtle May 29 '15 at 19:51
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    $\begingroup$ I assume your $A$ is the whole poset. The $\sup$ of $A$ exists; it's $\inf(\varnothing)$. Every element of $A$ is a lower bound for $\varnothing$ vacuously, and $\inf(\varnothing)$ is greatest among lower bounds, i.e. greatest among all elements. $\endgroup$ – anon May 29 '15 at 19:56
  • $\begingroup$ Yes, $(A,\le)$ is my poset. Because the inf always exists then $\inf(\emptyset)\in A$, and if there were another element, say $x$, greater than all elements of $A$ the above argument would also show that it's less than or equal to $\inf(\emptyset)$ hence equal to $\inf(\emptyset)$. I guess working with the empty set is a little weird, I'm actually surprised there was nothing wrong with the homework. $\endgroup$ – Squirtle May 29 '15 at 20:09

Here's a resolution: every subset has an inf, including the empty set. The inf of the empty set is a maximal element for the poset, so every subset has an upper bound.


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