Proof of the intersection of boundary of a set and a connected set is nonempty

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Use the relative topology of $C$: Let $A=int(E)\cap C$ and $B=int(E^c) \cap C$. Both $A$ and $B$ are open in $C$ and $A\cap B=\emptyset$ right? Now show that $A\cup B \neq C$ because $C$ is connected. Finally, show that $\partial E \cap C = C\backslash (A\cup B)$.

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Chans

Updated on December 15, 2020

Comments

  • Chans
    Chans almost 2 years

    Please help me to show that the intersection of a boundary of a set and a connected set(C) is not empty if the intersection of interior of E and C is not empty and the intersection interior of complement of E and C is not empty.

    • Chans
      Chans almost 9 years
      Trying to prove by contradiction. if C∩∂E=∅, then there exists a point x in C which can not be in ∂E or the other way around. But i don't know how to continue from there. Do you have any idea..
    • Sergio Parreiras
      Sergio Parreiras almost 9 years
      please see my answer
  • Chans
    Chans almost 9 years
    Hi Sergio I still couldn't finish it. I know A∩B=∅ , but I am still trying to prove that A∪B≠C.. Please give me one more step to continue the proof from here.
  • Chans
    Chans almost 9 years
    Hi Sergio, can we use the following to show that A∪B≠C. Since C is not a subset of A, c is not a subset of A∪B. Thus C≠A∪B.
  • Chans
    Chans almost 9 years
    Hi Sergio, can we use the following to show that A∪B≠C. Since C is not a subset of A, c is not a subset of A∪B. Thus C≠A∪B. By taking the intersection of both sides with ∂E we can show the latter, right?
  • Sergio Parreiras
    Sergio Parreiras almost 9 years
    @Chans: No you have to use the fact $C$ is connected. We have $A,B$ open, non-empty with $A\subset C$ and $B\subset C$, if we also have $A\cup B=C$ then $C$ would not be connected as you would be abe to split it in two disjoint, non-empty, open sets.