# Showing the Inclusion is sup-continuous

I fear I over simplified the following problem:

For any partially ordered set $(A,\leq)$, let $A^* = A- \{\max A,\min A\}$ if $\max A$ and $\min A$ exist. Show the inclusion $(A^*,\leq)\hookrightarrow (A,\leq)$ is $\sup$-continuous.

So I took any set $B\subseteq A^*$ such that $\sup B$ exists. Then $\iota(B)=B$, and $\iota(\sup B)=\sup B$. Then $\iota(\sup B)=\sup B=\sup(\iota(B))$, and so $\iota$ is $\sup$-continuous.

This seems too simple so I'm sure I've misinterpreted something. Can someone point out the source of error?

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If I've correctly hit on the point you were stumbling over, perhaps it would help to adjust notation to keep in mind that sups are conditional, e.g., $\sup_A (B)$ for the sup of $B$ in $A$.
Thanks, so I simply have to check for $maxA$ and $minA$. Am I correct in inferring that the intended result of the problem is not actually true? For $\iota$ to be sup-continuous $\iota(sup_{A^*}B)$ must equal $sup_{A}(\iota(B))$ for all $B\subseteq A^{*}$ But $\iota(sup_{A^*}B)\neq sup_{A}(\iota(B))$ when $B$ is empty as you showed in your example. – xdmf Sep 12 '10 at 4:39
Great, thank you, much appreciated. There is no mention anywhere in these notes that $B$ must be nonempty, so I take it that hypothesis is missing, as you said. If I were registered, I would vote up. – xdmf Sep 12 '10 at 4:54