# when inf and sup of a subset achievable in itself?

I was wondering if the following is true: In a topological space with partial order, the inf and sup for a closed subset are achievable inside the subset so that they become minimum and maximum.

Is it still true if I replace "a topological space with partial order" with "a topological space with total order"?

In a topological space, what other kinds of condition can make inf and sup of a subset achievable in itself?

Thanks and regards!

More question:

In an "ordered" (not sure what kinds of order is proper here) topological space, are inf and sup of a subset accumulation points of the subset?

More questions again:

In Euclidean space, are inf and sup for a closed subset inside the subset? Are they accumulation points of the subset? What if in metric space? Thanks!

-
If you have a metric on a linearly ordered topological space, the required property is called "completeness". –  user1119 Aug 17 '10 at 18:21

1 and 2: No. Take an interval $(\alpha, \beta)$ in $\mathbb{Q}$ where $\alpha, \beta$ are irrational. You can verify that such an interval is closed and that it contains neither its infimum nor its supremum. The property you want, for totally ordered sets, is called completeness.