Find a closed interval topology

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SUMMARY

The discussion centers on proving that an ordered set X, where every closed interval is compact, possesses the least upper bound property. This property asserts that every nonempty subset of X that is bounded from above has a least upper bound. The participants explore the implications of compactness, specifically that every open cover of a closed interval must have a finite subcover, leading to a contradiction if a greatest element does not exist in a bounded subset.

PREREQUISITES
  • Understanding of ordered sets and their properties
  • Knowledge of compactness in topology
  • Familiarity with the least upper bound property
  • Basic concepts of open and closed intervals
NEXT STEPS
  • Study the implications of compactness in metric spaces
  • Learn about the properties of ordered sets in topology
  • Research the relationship between closed intervals and compactness
  • Explore examples of sets that exhibit the least upper bound property
USEFUL FOR

Mathematics students, particularly those studying real analysis or topology, as well as educators looking for examples of compactness and the least upper bound property in ordered sets.

g1990
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Homework Statement


Let X be an ordered set where every closed interval is compact. Prove that X has the least upper bound property.


Homework Equations


X having the least upper bound property means that every nonempty subset that is bounded from above has a least upper bound, in other words, an upper bound where any number less that it is not an upper bound.
Compact means that every open cover has a finite subcover.


The Attempt at a Solution


Let A be a subset of X that is bounded from above. I know I should try to find a closed interval from this, but I'm not sure where to get it from.
 
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Hello,
I'll try to give an intuitive explanation.

"Let A be a subset of X that is bounded from above."
If A has a greatest element, we're done. If A does not, we can focus on an interval B of the form [...[ with A \cap B = B. Note that B has no greatest element thus there is an open cover that has no finite subcover (is there?).
Define C = \overline{B}.
  • If C = B, then B is closed and thus compact (CONTRADICTION! (?))
  • So C has at least an element more than B. What is this element?
 

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