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I am presently attempting Exercise 2 in Kaplansky, Set Theory and Metric Spaces

Exercise 2: Let $L$ be a partially ordered set in which every subset has a top and bottom element. Prove that $L$ is a finite chain.

Proof: Denote a subset of $L$ by $S$. If $S$ has a top and bottom element, then $\sup S$ and $\inf S$ exist and are elements of $S$. Denote them $a$ and $A$ respectively. Since $a,A \in S$ then $S$ is finite. This means that all the elements in $S$ can be ordered from smallest to largest, thus we have: $S= \{ a, ..., A \}$. Based upon this ordering, given any two elements, $b,c \in S$ one may determine that $b \le c$ or $c \le b$. Since all the subsets of L are a chain, this implies that L must also be a chain.

My Question: I am uncertain on whether the step in bold is valid. I believe it is based upon the transitivity condition in the definition for a partially ordered set, though I would appreciate some feedback on the matter.

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I don't see a reason given why $S$ is finite. –  Jonas Meyer Jun 6 '13 at 5:19
@user14111 Since the proof is that L is a chain, doesn't the upper and lower bounds of the empty set consist of the elements of L and thus would also have a top and bottom element? –  GovEcon Jun 6 '13 at 5:23
@Jordan: I don't understand your comment, but the answer is no. A top element of a subset would have to be an element of the subset, by the usual conventions of language. The empty set has no elements, so it cannot have a top or bottom element. –  Jonas Meyer Jun 6 '13 at 5:26
@CameronBuie Kaplansky defines a finite set informally by providing the example {1,2,...,99}. He later provides the function o(A) which gives the number of elements in set A. I took this as a sign that the definition for a finite set is when o(A) is an actually number. –  GovEcon Jun 6 '13 at 5:44
Thanks. I realized that it wasn't actually relevant how finiteness was defined in this case, though. –  Cameron Buie Jun 6 '13 at 5:50

2 Answers 2

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Given any two elements $x,y\in L,$ we know that $\{x,y\}$ has a top element and a bottom element. This shows that comparability holds on $L$, so since $L$ is a poset, then $L$ is a linearly ordered set.

Now, suppose that $L$ is not finite. Let $x_0$ be the top element of $L$, and for any nonnegative integer $n,$ let $x_{n+1}$ be the top element of $L\setminus\{x_0,...,x_n\}.$ Show that $\{x_n:n\text{ a nonnegative integer}\}$ is a subset of $L$ without a bottom element--the desired contradiction (you'll need the assumption that $L$ isn't finite).

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No, it is not valid. There are partially ordered sets, finite or infinite, that are not chains. (E.g., consider a power set of a set with at least $2$ elements, ordered by inclusion.) You have to use the hypothesis to show that $L$ is a chain. I recommend contraposition. You can show that if there are $2$ incomparable elements, then there exists a nonempty subset without a top and bottom.

You also gave no valid reason why $L$ is finite.

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