I think this is rather a simple question in order theory, but if someone could explain it step by step that would be really useful. If we have an arbitrary set, then let's denote the set of all finite subsets $F$, and the set of all cofinite subsets $C$.
I want to check whether $F,C$ are complete or chain complete, when they are ordered by inclusion.
Thanks.
If the set--say $X$--is itself finite, then $\mathcal F$ and $\mathcal C$ are simply the power set, which is trivially complete and chain-complete. Now, suppose $X$ is infinite.
Now, assuming that every infinite set has a countably infinite subset, can you think of a chain of finite sets that has no upper bound in the set $\mathcal F$? (We can't pull this off without some Choice principle, and I think this is the weakest one that will do the trick.)
Observe that $\bigl\{\{x\}:x\in X\bigr\}$ is a subset of $\mathcal F$. Does it have an upper bound in $\mathcal F$?
Similar (and closely related) counterexamples will show that $\mathcal C$ is neither complete, nor (with sufficient Choice) chain-complete.
Added: In the case that (chain-)completeness refers only to upper bounds and not lower bounds, then in fact $\mathcal C$ will be (chain-)complete. After all, $X\in\mathcal C,$ and for all $A\in\mathcal C$ we have $A\subseteq X.$ This shows that each collection of elements of $\mathcal C$ has an upper bound. Moreover, since for each $A\in\mathcal C$ there are only finitely-many $B\in\mathcal C$ such that $A\subseteq B,$ then there is readily a least upper bound for all such collections.