Every subset has first and last element -> set is finite Let $X$ be a partially ordered set, so that every non empty subset of $X$ has a first and a last element.
Show that $X$ is a finite set. 
And what if every subset only has a first element? 
Well, I proved it is a linear order, but now I'm stuck. 
Anyone ready to clear things up for me? Thanks a lot!

One may also give a more "direct" proof. Assume $X$ is infinite. We shall construct an strictly increasing/decreasing sequence which will contradict the existence of a last/first element.
Start with some element $x_1\in X$. At least one of the sets $\left\{ x>x_1 \right\}, \left\{ x<x_1 \right\}$ is infinite. Assume without loss of generality $\left\{ x>x_1 \right\}$ is infinite. By assumption, this set (as a subset of $X$), has a first element $x_2$. The set $\left\{ x>x_2 \right\}=\left\{ x>x_1 \right\}\setminus \left\{ x_2 \right\}$ is also infinite, and also contains a first element $x_3$. In this fashion we obtain a nested sequence of nonempty sets $\left\{ x>x_n \right\} \supsetneq \left\{ x>x_{n+1} \right\}$. Picking an element from each set which isn't the last one yields a strictly increasing sequence, which is a subset of $X$ without a last element. Contradiction.
(If you don't like the freedom of "picking an element", just use the sequence $ \left\{ x_n \right\}$.)
