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I am trying to prove that every neighborhood of a boundary bound contains a point in interior and $X \setminus A$ where $A$ is the set in consideration. I am given the following definitions

(1) $(X,\mathcal{T})$ is a topological space if $X,\emptyset \in \mathcal{T}$, any arbitrary union of open sets in $\mathcal{T}$ is an open set, and any finite intersection of an open set in $\mathcal{T}$ is an open set.

(2) A neighborhood of a subset of $X$ is an open set containing the subset.

(3) A set $A \subseteq X$ is closed if $X \setminus A$ is open

(4) $\bar A$ = $\cap\{ B \subseteq X : A \subseteq B \text{ and } B \text{ is closed in X} \}$

(5) Int($A$) = $\cup \{ B \subseteq A : B \text{ is open in } X \}$

(6) Ext($A$) = $X \setminus \bar A$

(7) $\partial A$ = $X \setminus ( \text{Int(}A)\text{)}\cup \text{Ext(}A \text{)})$

From this I can derive the closed definition of a topology and that $\partial A$ is closed. I am having trouble seeing how the previous facts and definitions give me that every neighborhood of a point in the boundary contains a point in $A$ and $X \setminus A$ where $X$ is the space and $A$ is some subset of the space. What are some derivations that would help me figure this out?


What I think it comes down to is showing that given an open neighborhood $\mathcal{O}$ of some point $x \in \partial A$, $\mathcal{O} \cap Int(A)\ne \emptyset$ and $\mathcal{O} \cap (X \setminus A) \ne \emptyset$

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  • $\begingroup$ You only need to show that the neighborhood contains a point of $X\setminus A$ not $X\setminus\overline{A}$. For this consider the case where $X=\overline{A}$ say for example $X=[0, 1]$ and $A=(0, 1)$. $\endgroup$ – Owen Sizemore Sep 5 '13 at 21:14
  • $\begingroup$ I can see it with specific cases, such as with the standard metric topology, or with a discrete space, but in the general sense, it does not make much sense to me. $\endgroup$ – user1876508 Sep 5 '13 at 21:18
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Using the definitions straightforwardly:

Let $x\in\partial A$, and let $B$ be a neighbourhood of $x$. Suppose that $A\cap B=\emptyset$. This implies that $x$ is an interior point of $X\setminus A$: indeed, $A\subset X\setminus B$, which is a closed set; as $x\not\in X\setminus B$, this shows that $x\not\in\bar A$. But this would imply $x\in\mbox{Ext}(A)$ and $x\not\in\partial A$. This contradiction shows that $A\cap B\ne\emptyset$.

If $(X\setminus \bar A)\cap B=\emptyset$, then $y\not\in \mbox{Ext}(A)$ for every $y\in B$. This implies that $B\subset A$, which makes $x$ an interior point of $A$. But this contradicts the fact that $x\in \partial A$. We conclude that $(X\setminus\bar A)\cap B\ne\emptyset$.

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Observe that if $x$ is in the boundary of $A$, then $x$ is not in the interior or the exterior of $A$. Since $x\notin\operatorname{int}A$, $x$ has no open nbhd contained entirely in $A$. But that just means that if $U$ is an open nbhd of $x$, then $U\nsubseteq A$, and hence $U\cap(X\setminus A)\ne\varnothing$. A similar argument shows that since $x$ is not in the exterior of $A$, every open nbhd of $x$ must meet $A$ in a non-empty set. The converse is equally easy to show: if every open nbhd of $x$ meets both $A$ and $X\setminus A$, then $x$ is in the boundary of $A$. To complete the argument, prove the following proposition:

Proposition. For any set $S\subseteq X$ and any $x\in X$, $x\in\operatorname{cl}S$ if and only if $U\cap S\ne\varnothing$ for each open nbhd $U$ of $x$.

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  • $\begingroup$ Why must $U \cap (X \setminus A) \ne \emptyset$? It sounds like you are implicitly saying that there is no open set in the boundary, but that fact I cannot see. $\endgroup$ – user1876508 Sep 5 '13 at 22:07
  • $\begingroup$ @user1876508: Because $U\nsubseteq A$: $U$ ‘sticks out of’ $A$, so it must contain some point that’s not in $A$. $\endgroup$ – Brian M. Scott Sep 5 '13 at 22:08
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    $\begingroup$ I understand that, but my grader would not be happy if I said that it just sticks out, I think the hidden message is that there cannot be an open set in the boundary. $\endgroup$ – user1876508 Sep 5 '13 at 22:09
  • $\begingroup$ @user1876508: If your grader cannot follow the step from $U\nsubseteq A$ to $U\cap(X\setminus A)\ne\varnothing$, then he or she is incompetent: it’s an utterly trivial step that requires no further justification. $\endgroup$ – Brian M. Scott Sep 5 '13 at 22:13
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    $\begingroup$ Why is it trivial? If it was trivial, then I would be able to say that it is crystal clear. From the definitions that I have been given, it is not trivial for me to see that what you have shown is true. $\endgroup$ – user1876508 Sep 5 '13 at 22:14

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