Real Analysis: Interior, Closure and Boundary

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SUMMARY

The discussion focuses on proving the equivalence of three conditions regarding a subset W of S in the context of real analysis. Specifically, it establishes that W is relatively closed in S if and only if W equals the intersection of its closure with S, and the boundary of W intersected with S is a subset of W. The key definitions utilized include those of open and closed sets, as well as the concepts of boundary, interior, and closure. Participants are encouraged to explore the implications of these conditions through logical deductions and diagrammatic representations.

PREREQUISITES
  • Understanding of open and closed sets in topology
  • Familiarity with the concepts of boundary, interior, and closure in real analysis
  • Knowledge of the properties of relative openness and closedness
  • Ability to manipulate set operations and intersections
NEXT STEPS
  • Study the definitions and properties of closure and interior in metric spaces
  • Learn about the implications of boundary conditions in topology
  • Explore examples of relatively closed sets in various topological spaces
  • Practice proving set equivalences using diagrammatic methods
USEFUL FOR

Students of real analysis, mathematicians focusing on topology, and anyone interested in the properties of sets in metric spaces will benefit from this discussion.

michael.wes
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Homework Statement


Let [tex]W\subset S \subset \mathbb{R}^n.[/tex] Show that the following are equivalent: (i) [tex]W[/tex] is relatively closed in [tex]S[/tex], (ii) [tex]W = \bar{W}\cap S[/tex] and (iii) [tex](\partial W)\cap S \subset W[/tex].

Homework Equations


The only thing we have to work with is the definitions of open and closed sets, relatively open and relatively closed sets, the result that the complement of an open set is closed, the definition of boundary, interior and closure.

The Attempt at a Solution


I have proved that (ii) implies (i). I need help with (i) implies (iii) and (iii) implies (ii).

For (i) implies (iii), there exists a closed set [tex]C[/tex] such that [tex]W=C\cap S[/tex]. Now try to derive that [tex]\partial W \cap S \subset W[/tex]. Write [tex]\partial(C\cap S)\cap S[/tex], but after this all I am able to do is draw a diagram and get stuck.

For (iii) implies (ii), we have that [tex]\partial W \cap S \subset W[/tex] and we need to derive [tex]W = (W\cup\partial W) \cap S[/tex], but I can't get started.

Please help. Thank you!
 
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Try the direction (i) => (ii) => (iii) => (i) instead. You will need the relation [tex]\overline{W} = W^\circ \cup \partial W[/tex].
 

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