Let $M$ be an ordered set such that
$a)$ $M$ has a maximum element
$b)$ Every (non-empty) subset of $M$ has an infimun.
Prove that every non-empty subset of $M$ also has a supremum.
I tried fixing a subset $A$ and considering its set of majorants, which is non-empty (since it contains $\sup M$), so must have an infimum $a'$. What is left to prove is that $a'$ belongs to this set, but I cannnot follow this up. Any hints are appreciated.
Source (Romanian book): Ion D. Ion, R. Nicolae, "Algebra", Editura Didactica si Pedagogica, Bucharest, $1981$ – problem $9$, page $22$.
Best Answer
Let's call $U(X)$ the set of upper bounds of $X$ and $L(X)$ the set of lower bounds of $X$. It is generally true that $L(U(X))\supseteq X$. By definition, $\inf X=\max L(X)$, and therefore $\inf U(A)=\max L(U(A))$. Since $L(U(A))\supseteq A$, we have that $\inf U(A)$ is larger than all the elements of $A$, and therefore $\inf U(A)\in U(A)$.