Fie \( X \) un spatiu arc-conex si \( A \) si \( B \) doua submultimi ale sale disjuncte si nevide.
Daca bordul lui \( A \) = bordul lui \( B \) rezulta ca \( \overline{A}\cup B=X \) ?
AMM 3011 - Calin Popescu
Spatiu arc conex
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Bordul lui \( A \) e cumva \( X\setminus (int(A)\cup ext(A)) \)?
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Pai cred ca daca am inteles bine problema e foarte simpla. \( \mathbb{R} \) e arc-conex. Daca luam \( A=[0,1]\cap \mathbb{Q}, B=[0,1]\cap (\mathbb{R}\setminus \mathbb{Q}) \), atunci \( A,B \) sunt disjuncte si \( bd(A)=bd(B)=[0,1] \). Dar \( \bar{A}\cup B=[0,1]\neq \mathbb{R} \).
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In ceea ce am scris \( ext(A) \) este exteriorul lui \( A \) sau interiorul complementarei lui \( A \).Dragos Fratila wrote:Daca ext inseamna elementele din inchidere care nu sunt in multime atunci da.
Se mai poate scrie \( \partial(A) = \overline{A}-A^\circ \), unde \( A^\circ \) este interiorul lui A.
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