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.


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.
  • 3
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Nov 4 '14 at 4:11
  • 1
    $\begingroup$ @Asaf: But the easiest way to show that it’s countable is by contradiction. $\endgroup$ – Brian M. Scott Nov 4 '14 at 4:12
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Nov 4 '14 at 11:17

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