If every nonempty subset of a set $S$ has a least and greatest element, is $S$ finite?

1.6k Views Asked by At

Some sets are well-ordered; all of their nonempty subsets have least elements. You can also have sets where all of their nonempty subsets have greatest elements.

Some sets have both of these properties. For example, any totally-ordered finite set will satisfy both of the above properties.

Are there any infinite sets that have both of these properties? I can't think of any off the top of my head, but I also can't seem to find a proof that says they can't exist.

Thanks!

2

There are 2 best solutions below

0
On BEST ANSWER

$S$ must indeed be finite. Suppose that $S$ is infinite. Let $S_0=S$ and $x_0=\max S_0$. Given $S_n\ne\varnothing$ for some $n\in\omega$, let $x_n=\max S_n$, and let $S_{n+1}=S_n\setminus\{x_n\}$. Then the set $\{x_n:n\in\omega\}$ has no least element.

0
On

The whole set $S$ has a least element $a_1$ and a greatest element $b_1$.

The set $A\setminus\{a_1,b_1\}$ has a least element $a_2$ and a greatest element $b_2$.

The set $A\setminus\{a_1,a_2,b_1,b_2\}$ has [etc.etc.]

The subset $\{a_1,a_2,a_3,\ldots\}$ has no greatest element unless it's finite; similarly $\{b_1,b_2,b_3,\ldots\}$ has no least element unless it's finite.