# A question about cofinal totally ordered sets.

Let $A$ be an uncountable set, and let $L$ be the poset consisting of all finite subsets of $A$ (the ordering on $L$ is inclusion). Show that $L$ does not have a totally ordered cofinal subset.

I am not really sure about the best way to go at this. I am not sure how to deal with uncountable sets; this is new to me. Thanks in advance.

## 1 Answer

HINT: Suppose that $C$ is linearly ordered and cofinal in $L$.

• Show that since $C$ is cofinal in $L$, it must be uncountable.
• Show that there is some $n\in\Bbb Z^+$ such that uncountably many members of $C$ have cardinality $n$.
• Derive a contradiction.
• There's really no need for contradiction here. Pick a linearly ordered set, show it's countable, conclude that its union is countable, therefore it's not cofinal. – Asaf Karagila Nov 4 '14 at 4:11
• @Asaf: But the easiest way to show that it’s countable is by contradiction. – Brian M. Scott Nov 4 '14 at 4:12
• Comments are not for extended discussion; this conversation has been moved to chat. – user642796 Nov 4 '14 at 11:17