1

I have the following task:

Show that in topological space $\partial\partial A \subset \partial A$, where $\partial A$ denotes the set of boundary points of $A$, that is $\partial A = \overline{A}\;\setminus\;A^o$, where $\overline{A}$ is the closure of $A$ and $A^o$ is the set of interior points of $A$. Give an example of $A$ such that $\partial \partial A \neq \partial A$.

I tried to solve this but I resulted in equality $\partial \partial A = \partial A$, I'm not sure where I went wrong. Here's what I tried:

By the definition of $\partial A$ we have $\partial A\;\cup A^o = \overline{A}$, that is $A^o$ is the open complement of $\partial A$ in $\overline{A}$. Now

$$\partial \partial A = \overline{\partial A}\;\setminus\; (\partial A)^o = \overline{\partial A}.$$

Now because $\partial A$ is closed in $\overline{A}$ then $\partial A$ is the smallest closed set containing $\partial A$, that is $\overline{\partial A}=\partial A$, so

$$\partial \partial A = \overline{\partial A}=\partial A,$$

that is $\partial \partial A = \partial A$

I know I did a mistake in some point...

jjepsuomi
  • 8,619
  • 2
    Could you show us your proof? – AnalysisStudent0414 Jan 25 '16 at 19:01
  • Yes I can, it's not correct then but I'll add it anyway :) – jjepsuomi Jan 25 '16 at 19:02
  • $A = \mathbb{Q}$ shows that equality does not hold in general. –  Jan 25 '16 at 19:02
  • NB The above duplicate also gives an easy way to construct a set with proper inclusion: Just look for a set whose boundary has nonempty interior, such as @menag's example. –  Jan 25 '16 at 19:06
  • The boundary of a set is always closed, hence equal to its closure, which gives the containment you want. However, the boundary of a set does not always have empty interior. If it does have nonempty interior, then the second boundary operation removes points, which gives strict containment. – Ian Jan 25 '16 at 19:10
  • Thank you @Ian I see now my mistake. – jjepsuomi Jan 25 '16 at 19:13
  • P.S. should I close this question, if it's duplicate.. – jjepsuomi Jan 25 '16 at 19:24

0 Answers0