The question is really in the title. I would like to prove that there exists the minimum for of a partially ordered set $(P,\leq)$ having the property that any totally ordered subset (chain) has minimum, and that for any two elements in $P$ (and hence for any finite subset) there exists in $P$ a lower bound for these elements.
In every counterexample that I've tried to define always comes up in the end a chain not having minumum, so I persuaded myself that this should be true, but I don't know how to prove it, since not every poset is finite union of chains... Does anyone have suggestions about that?
Thank you so much!
Suppose $P$ not having a minimum. Let's define a chain not having minimum. By induction, set $x_0$ any element in $P$, since $x_0$ is no minimum, there exists $y$ such that $y\not\geq x_0$, that means either $y<x_0$ or $y$ and $x_0$ are not comparable.
In the former case set $x_1=y$, in the latter find an element $z\leq x_0,y$ (which won't be neither of the two, o/w they would be comparable) and set $x_1=z$.