Let $C$ be a nonempty compact subset of $R^n$, with a certain partial order defined on it. I am trying to prove that $C$ contains a maximal element.
My idea is: start with a certain element of $C$. If it is not maximal, replace it with a larger element, and so on. This process results in a sequence of elements in $C$. Because of compactness, this sequence has a subsequence that converges to an element in $C$. This limit, $l$, must be a maximal element.
Is this true? I am particularly unsure about the last statement - is this true that the limit must be a maximal element?
ADDITION: If the claim is not true, what condition can be added in order to make it true?
A second question is: if this theorem (possibly with an additional condition) is indeed true, is it a well-known fact that I can cite in a paper without having to prove it?
Counterexample: Consider the compact subset $[0,1] \subset \mathbb{R}$. Define the usual order on $[0,1) \subset [0,1]$, and extend to $[0,1]$ by defining $ 1 \leq x$ for all $x \in [0,1)$.
A couple of observations: (1) Given an ascending sequence approaching $1$, its limit (i.e. 1) is not maximal. (2) We could use your approach to find an ascending sequence that has limit $1/2$, say $x_n= \frac{1}{2} - \frac{1}{n}$. But $\frac{1}{2}$ is clearly not maximal in $[0,1]$, even with the usual ordering.
How to salvage it: Since you want to use the compactness of the set, it makes sense to at least have the partial order respect limits. That is, if $x_n \leq x_{n+1}$ for all $n$ and $x_n \to a$, then $x_n \leq a$ for all $n$.