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!
 A: Since $X$ is a linear order and every subset has a least element, $X$ is a well-order. It follows that it is order-isomorphic to some ordinal $\alpha$, so it is equivalent to consider the same question for $\alpha$.
$\omega$ has no maximal element, hence $\omega \nsubseteq \alpha \implies \omega \notin \alpha \land \omega \neq \alpha$. It follows that $\alpha \in \omega$. Hence $|\alpha| \leq \alpha < \aleph_0$.
A: If it's a linear ordering with a first and last element, then it's in bijection with $\{1,...,n\}$ for some $n$. Build the bijection inductively, and you'll be done. There's a first element that should definitely map to $1$. Keep going, and eventually you'll have to stop because there's a last element. Wherever you stopped is the $n$ you want.
To answer your other question, any ordinal (with it's natural ordering) satisfies that every subset has a least element, as it's a well-ordering. In particular, the naturals numbers have this property.
A: 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\}$.)
