0
$\begingroup$

From Wikipedia https://en.wikipedia.org/wiki/Completeness_(order_theory)#Relationships_between_completeness_properties

"the existence of all suprema ... equivalent to the existence of all infima. Indeed, for any subset $X$ of a poset, one can consider its set of lower bounds $B$. The supremum of $B$ is then equal to the infimum of $X$"

My question is, can $B$ be the empty set?

My idea is that $B$ cannot be the empty. This is because if we consider the empty set, then all elements of the poset is an upper bound of the empty set. But the poset has no lower bound because $X$ has no lower bound, so there is no least upper bound on the empty set, contradicting the existence of all suprema. Is this argument sound?

$\endgroup$
  • 2
    $\begingroup$ You're right. The paragraph needs a few more details to be considered a complete proof. However, it quite clearly illustrates the idea that a full proof might work. $\endgroup$ – Arthur Feb 17 '17 at 8:08
1
$\begingroup$

The trick here is that if there exists a supremum of $X$ for every subset $X$ of a poset $P$, then there exists the supremum of the empty set in $P$, that is, $\bigvee\varnothing \in P$.
It follows that $\bigvee\varnothing = \bigwedge P$, and this covers the case that you rose, because if $X$ had no lower bound, then $\bigwedge X = \bigvee\varnothing \in P$, a contradiction.
So this means that if there is a subset with no lower bounds, then there must be some subset without supremum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.