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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!

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If you have a metric on a linearly ordered topological space, the required property is called "completeness". –  user1119 Aug 17 '10 at 18:21
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1 Answer 1

up vote 4 down vote accepted

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.

3: I will assume you mean a totally ordered set with its order topology. In that case, this is not always true if the infimum or supremum are contained in the subset but true otherwise; this is just a matter of working through the definitions.

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Thanks! For 3, why sup and inf are not accumulation points when they are in the subset? –  Tim Aug 17 '10 at 18:43
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Just take points that are isolated; e.g., in R, take {0,1}. Sup is 1, but is not an accumulation point; the inf is 0, but is not an accumulation point. –  Arturo Magidin Aug 17 '10 at 18:46
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Sorry, instead of "false" I meant to say "not always true." Consider for example the infimum and supremum of a closed interval [a, b] in Z. –  Qiaochu Yuan Aug 17 '10 at 18:46
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@Tim: in R, the infimum and supremum of a closed subset, when they exist, are in that subset. This is a consequence of the fact that R is complete (in both the order-theoretic and metric senses). –  Qiaochu Yuan Aug 18 '10 at 0:02
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@Tim: I have changed the link in the answer to the notion of completeness that is relevant to totally ordered sets. Note that you need some form of completeness to ensure that infima and suprema even exist. –  Qiaochu Yuan Aug 18 '10 at 0:32
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