I understand the boundary to mean all the points that are in the set AND not in the set. For example in $\mathbb{R}^2$. $S=\{(x,y)\; | \;(x^2)-x\leq y \leq 0\}$. This is a parabola that opens upward. The $bd(S)=\{(x,y)|(x^2)-x= and y \leq 0\} \cup \{(x,0)\; | \;0\leq x\leq 1\}$. I'm not seeing how this is related to writing it in terms of $bd(S)= Cl(a)\cap Cl(A^c)$. maybe someone could help me understand $cl(S^c)$ because $cl(S)=S$ and I do not see how it would equal the boundary listed above

  • $\begingroup$ Please let me know if I edited your sets above correctly. $\endgroup$ – Mnifldz Feb 23 '15 at 23:11
  • $\begingroup$ I think you mean all the points that are in the closure of the set and not in the interior of the set. $\endgroup$ – Uncountable Feb 24 '15 at 0:15
  • $\begingroup$ does that mean cl(S^c)=int(S) ? @Uncountable $\endgroup$ – Nicole Feb 24 '15 at 0:33
  • $\begingroup$ I was commenting on your definition of the boundary of a set, which seems contradictory to me. $\endgroup$ – Uncountable Feb 24 '15 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.