let $X$ be a topological space. Let $\emptyset \neq A \subset X$ be such that $A=\partial A$

Which one of the following is then guaranteed to be true?

i) $\exists x \in A: \forall N \in N(x): N \cap (X \backslash A)=\emptyset$

ii) $int(A) \neq \emptyset$

iii) $A= \bar{A} $ Where $\bar{A}$ is the contact point which our teacher denoted boundary points + interior points

My own thoughts to this question are the following:

it cannot be ii) since $int(A)$ must be empty since $A$ only contains the boundary points.

Furthermore it cannot be iii) because $\bar{A}$ contains interior points as well where $\partial A=A$ only contains boundary points

When I try to paint i) I cannot make it fit as well. My thoughts are that since the boundary points of A must be contained in X since $A \subset X$. Then when I look at the last $N \cap (X \backslash A)=\emptyset$ I cannot see how the intersection between the neighborhoods without $A$ is empty. I mean if you remove the set the neighborhoods of the set will still be there? And when the Neighborhoods are still there they must still be contained in X? So in my view it would make more since if it was $N \cap (X \backslash A) \neq \emptyset$. But I could be misunderstanding something.

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    $\begingroup$ iii) is correct because $\overline A=\partial A\cup\operatorname{int}(A)=A\cup\emptyset=A$. -- For i), consider $X=\Bbb R$ and $A=\{0\}$ $\endgroup$ – Hagen von Eitzen Jun 3 '18 at 20:33
  • $\begingroup$ In number 1, does $N(x)$ mean the set of all neighborhoods of $x$. I've never seen that notation. $\endgroup$ – fleablood Jun 3 '18 at 20:36
  • $\begingroup$ Thank you Hagen I see it now since $\hat{A}=int(A) \cup \partial A$ and $int(A)$ is empty we're just left with the union of the empty set and the boundary points of A, which is just the boundary points of A. Would i) be equal if we had $\neq$ instead of = $\endgroup$ – user420309 Jun 3 '18 at 20:40
  • $\begingroup$ Fleabodd: yes it should be read as : there exists an x in A such that for all neighborhoods around x $\endgroup$ – user420309 Jun 3 '18 at 20:42
  • $\begingroup$ Isn't a boundary the set of points that are limit poits of A and X/A? So be definition every neighborhood of a boundary point with have points that are in A (unless A has not limit points) and points that are in X/A (unless X/A has no limit points). If X/A has no limit points the i) is possible but in general if X/A does have limit points i is not possible. $\endgroup$ – fleablood Jun 3 '18 at 20:43

The boundary of any set is closed, as $\partial A = \overline{A} \cap \overline{X\setminus A}$ is the intersection of closed sets.

So if $A = \partial A$, $A$ is closed and so $\overline{A} = A$.


Number $i$ isn't correct for example suppose that $X = R^2$ that $R$ is a real space and $A = S^1$ for any $x$ in $S^1$, any neighborhood of $x$, $N \cap (X - A)$ isn't empty. Number $ii$ is wrong and $iii$ is correct.


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