Proving the Interior of a Boundary for Open Sets

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SUMMARY

The discussion centers on proving that the interior of the union of an open set U and its boundary, denoted as int(U ∪ Bdy(U)), equals the interior of U, or Int(U), when U is an open set. The boundary of U is defined as Bdy(U) = closure(U) ∩ closure(X - U). The participant attempts to demonstrate that any point in the boundary of U has neighborhoods that intersect the complement of U, ultimately leading to the conclusion that the assertion is false when U is the union of the open intervals (0,1) and (1,2).

PREREQUISITES
  • Understanding of open sets in topology
  • Familiarity with the concepts of closure and boundary in topological spaces
  • Knowledge of neighborhoods and their properties
  • Basic proficiency in set theory and mathematical proofs
NEXT STEPS
  • Study the properties of open sets and their interiors in topology
  • Explore the definitions and implications of closure and boundary in topological spaces
  • Learn about counterexamples in topology to understand common misconceptions
  • Investigate the relationship between open sets and their complements in various topological contexts
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Mathematics students, particularly those studying topology, and educators seeking to clarify concepts related to open sets, boundaries, and interiors in mathematical proofs.

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Homework Statement


I need to prove that the int(U union Bdy(U))=Int(U) when U is open.


Homework Equations


Bdy(U)=closure(U) intersect closure(X-U)
a point is in the interior if there is an open neighborhood of the point that is contained in the set.


The Attempt at a Solution


obviously, if x is in U and U is open, there is a neighborhood of x in U by the fact that U is open. If x is in Bdy(U), then I want to prove that every open neighborhood of x is not in S=U union Bdy(U). Now, by the definition of boundary, every open neighborhood of x intersects X-U, so obviously there are points outside of U. What I need to prove is that there is a point y in every open neighborhood of x that is not in U AND is not in the closure of U, meaning that it is not in the boundary.
 
Physics news on Phys.org
Take U to be the union of the open intervals (0,1) and (1,2). What you are trying to prove is false.
 

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