If $L$ is a chain, prove it is finite. I need ideas on the following problem. Suppose $L$ is a poset and every subset $S$ of $L$ has a top and bottom element. Prove $L$ is a finite chain.
All I need to do is prove that $L$ is finite (I have already proved $L$ is a chain). Any ideas or suggestions on solving this problem would be great! Thanks.
 A: You've proved that $L$ is a chain, so since every non-empty subset of $L$ has a bottom element, then $L$ is well-ordered.
Thus, $L$ is order-isomorphic to a unique ordinal, say $\alpha$. Can you show that $\alpha$ must be finite? (Hint: Use the fact that every subset of $L$ has a top element.)

Alternate approach (related to my answer to this question, asked previously by another user): Let $x_0$ be the least element of $L,$ $x_1$ the second least element of $L,$ and so on. This sequence is well-defined recursively, because every non-empty subset of $L$ has a bottom element--so $x_1$ is the bottom element of the subset of $L$ containing all but $x_0$, $x_2$ is the bottom element of the subset of $L$ containing all but $x_0$ and $x_1,$ etc.--and at some point, this process must stop (namely, when we reach the top element of $L$) after finitely-many steps. If not, then the set of all $x_n$ (with $n$ a nonnegative integer) is a subset of $L$ without a top element! (Why?) This contradicts our assumption. Thus, our process will stop after finitely-many steps (say at $x_n$), giving us an increasing finite sequence $x_0,x_1,...,x_n$ of points of $L$. Show that every point of $L$ is in this sequence (use the recursive definition), and you're done.
