Consider a partially ordered set $A\subseteq \mathbb{R}^k$. I know that if $A$ is compact then $\sup A=\max A$. Could you give an intuition of why having $A$ closed is not enough to conclude $\sup A=\max A$? Assume that I allow the supremum to take an infinite value.
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This question and this question might help you if you haven't already seen them. – Noble Mushtak Apr 22 '16 at 11:03
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What do those terms even mean for a set without an order? – Tobias Kildetoft Apr 22 '16 at 11:03
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@TobiasKildetoft Ok thank you, I have added "partially ordered". – Star Apr 22 '16 at 11:09
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@NobleMushtak I still don't get why closeness is not enough – Star Apr 22 '16 at 11:10
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1With what order? – Tobias Kildetoft Apr 22 '16 at 11:10
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@TobiasKildetoft Thanks. Is specifying the order necessary to answer my question? Why? – Star Apr 22 '16 at 11:11
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1Because the existence of either sup or max depends on the ordering. – Tobias Kildetoft Apr 22 '16 at 11:11
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Ok, thank you. If possible, could you suggest an ordering that would guarantee the existence of the sup? – Star Apr 22 '16 at 11:20
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1Sure, pick the reverse of a well-order. Then any non-empty set will have a largest element, which will also be the sup. This will of course be a completely weird ordering. – Tobias Kildetoft Apr 22 '16 at 11:22
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$\sup(A)$ need not exist. If $A$ has a $\sup(A)$ and $\sup(A)$ is actually a member of the ordered set (so infinity(as a point not in the set above all points) is not allowed, because infinity can never be a maximum!) and $A$ is closed in the order topology, then $\sup(A) \in A$ and so $\sup(A) = \max(A)$. – Henno Brandsma Apr 22 '16 at 12:01
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Suppose $(X,<)$ is a linear order in the order topology and for some $A\subseteq X$, $A$ is closed and $\alpha=\sup(A)$ exists but it is not an element of $A$. Then, for every open interval $I$ containing $\alpha$ then by definition of supremum, there is an element $a\in A$ such that $a\in I$.
This shows that $\alpha$ is a limit point of $A$, and since $A$ is closed it should contain all its limit points, so $\alpha\in A$, and $\sup A=\max A$. I believe then that when considering the order topology it is the case that if $A$ is closed and $\sup A$ exists, $\sup A=\max A$. It could happen however that a closed set does not have maximum nor supremum, as for instance $A=[0,\infty)$ in the reals.
However, with a change of topology we might have different things. Consider the topology on $\mathbb{R}$ whose basic opens are sets of the form $[a,b)$. In this case, the set $A=(-\infty,0)$ is closed, does not have a maximum and $\sup A=0$ exists.
Darío G
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