Does a bounded well-ordered set have a maximum?

Question

Suppose $(A,\preccurlyeq)$ is a well-ordered chain, i.e., every subset $B\subseteq A$ has a least element under the order $\preccurlyeq$.

We say that a set $B\subseteq A$ is bounded if there exists $s\in A$ such that $b\preccurlyeq s$ for all $b\in B$.

Question: Does every bounded subset of $A$ have a maximum?

If $B$ is finite this is obviously true, since we can start with the minimum $m=\min B$, and keep applying the successor operation $$x^+=\min\textstyle_{\preccurlyeq}(B\smallsetminus\{b\in B: b\preccurlyeq x\})$$ and stop when we end up with a singleton $\{M\}$, this will be the maximum of $B$. But what if $B$ is infinite?

I can't think of an easy counterexample to this, it feels like it should be true.

Have you considered $A = {-\frac{1}{n} \mid n \in \mathbb{N}_+} \cup {0}$ and $B = A \setminus {0}$? — Dan Rust, Nov 20 '20 at 16:12
@DanRust No I have not Do you want to type it up as an answer? — Luke Collins, Nov 20 '20 at 16:15

score 1 · Accepted Answer · answered Nov 20 '20 at 16:19

1

If we let $A = \{-\frac{1}{n} \mid n \in \mathbb{N}_+\} \cup \{0\}$ with the usual order, then $A$ is well-ordered. Now take $B = A \setminus\{0\}$ which is bounded, as $0$ is an upper bound for $B$, but as $0$ is not an element of $B$, and no other element of $A$ provides an upper bound for $B$, it follows that $B$ has no maximum.

answered Nov 20 '20 at 16:19

Dan Rust

30,108

Thanks for this! I actually edited my question a bit, could you take a look? – Luke Collins Nov 20 '20 at 16:28
I feel like you should really be making this a new question, as your edit completely changes the content of your original question. – Dan Rust Nov 20 '20 at 16:37
you are right, I will do that. – Luke Collins Nov 20 '20 at 16:37

Does a bounded well-ordered set have a maximum?

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