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let $A \subseteq \mathbb{R}^n$. Let $X = \{ x \in \mathbb{R}^n : \forall \epsilon > 0, \; \; B(x, \epsilon) \cap A \neq \varnothing \; \; and \; \; B(x, \epsilon) \cap ( \mathbb{R}^n \setminus A ) \neq \varnothing \} $. I want to show that $\partial A = X$.

My attempt

Let $x \in \partial A$ be arbitrary. Let $\epsilon > 0 $.By definition, we can take a neighborhood $N$ of $x$ such that $N \cap A \neq \varnothing $ and $N \cap (\mathbb{R}^n \setminus A) \neq \varnothing $. Let $N = B(x, \epsilon) $. Hence, $x \in X$. So $\partial A \subseteq X$.

IF $x \in X$, then obviously $x \in \partial A$ by definition.

It seems that this problem is obvious. Maybe I am wrong? Can someone give me feedback to my solution? thanks.

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    $\begingroup$ What definition of $\partial A$ are you using? $\endgroup$ – dani_s Feb 8 '14 at 23:47
  • $\begingroup$ How do you define the boundary? The definition I'm familiar with is precisely the set of points every neighborhood of which intersects both the exterior and the interior of a set. $\endgroup$ – user99680 Feb 8 '14 at 23:47
  • $\begingroup$ yes this definition $\endgroup$ – user124140 Feb 8 '14 at 23:49
  • $\begingroup$ So I'm thinking OP's definition involved neighborhoods, and the problem is to show the same holds for this specific type of neighborhood (open balls $B(x,\epsilon)$) $\endgroup$ – MPW Feb 9 '14 at 0:42
  • $\begingroup$ @user99680: That's not right. It need only meet the set and its complement. Your definition would exclude isolated points, which are certainly part of the boundary. $\endgroup$ – MPW Feb 13 '14 at 5:14
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In the first part of your attempt you have mixed up "$\>\exists$ a neighborhood" and "$\>\forall$ neighborhoods". In the second part you have written "obviously" for something that is not obvious.

The essential point of this exercise is that the $B(x,\epsilon)$ form a neighborhood base at $x$, for each $x\in{\mathbb R}^n$.

Definition 1: Given a nonempty set $A\subset{\mathbb R}^n$, the boundary $\partial A$ consists of all points $x\in{\mathbb R}^n$ such that any neighborhood $N$ of $x$ intersects both $A$ and ${\mathbb R}^n\setminus A$.

Definition 2: Your set $X$.

Claim: $\partial A=X$.

Proof. Let $x\in\partial A$. Each ball $B(x,\epsilon)$, $\>\epsilon>0$, is a neighborhood of $x$ and therefore intersects both $A$ and ${\mathbb R}^n\setminus A$. It follows that $x\in X$, which proves $\partial A\subset X$.

Conversely, let $x\in X$, and consider an arbitrary neighborhood $N$ of $x$. By definition $N$ contains a ball $B(x,\epsilon)$ for some $\epsilon>0$. By assumption this ball intersects both $A$ and ${\mathbb R}^n\setminus A$, and so does $N$. It follows that $x\in\partial A$, which proves $X\subset\partial A$.

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It is kind of correct, you should have said that by definition ANY neighborhood $N$ of $x$ intersects both $A$ and $\mathbb R \setminus A$. Since $B(x, \epsilon)$ is a neighborhood of $x$ for every $\epsilon > 0$, it follows trivially that $x \in X$.

Proving $X \subset \partial A$ is actually harder: you need to prove that if any ball centered at $x$ intersects both $A$ and $\mathbb R \setminus A$, then so does any neighborhood of $x$.

EDIT: I guess I'll conclude the proof... Let $x \in X$ and let $N$ be a neighborhood of $x$; then, by definition of neighborhood, $N$ contains an open set $U$ containing $x$ and, by definition of open set, $U$ contains an open ball $B(x, \epsilon)$ for some $\epsilon > 0$. Thus we have $B(x, \epsilon) \subset N$, implying $A \cap B(x, \epsilon) \subset A \cap N$.

By hypothesis we have $B(x, \epsilon) \cap A \neq \emptyset$ and therefore $A \cap N \supset A \cap B(x, \epsilon) \neq \emptyset$, i.e. $A \cap N \neq \emptyset$. Similarly we also have $(\mathbb R^n \setminus A) \cap N \neq \emptyset$, which concludes the proof.

(To be clear, I'm using wikipedia's definition of neighborhood.)

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  • $\begingroup$ how can I show the other direction ? $\endgroup$ – user124140 Feb 9 '14 at 0:52
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    $\begingroup$ @Learner If $N$ is a neighborhood of $x$, by definition it contains an open set $U$ containing $x$, and again by definition $U$ contains a ball $B(x, \epsilon)$ for some $\epsilon > 0$... then? :) $\endgroup$ – dani_s Feb 9 '14 at 0:56
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Your definition of a boundary point of $A$ is that every neighborhood of such a point meets both $A$ and its complement.

The problem you have is to show that this is the same as requiring that every open ball about such a point meets both $A$ and its complement.

But this is immediate since the set of all open balls is a basis for the topology of the space. Every neighborhood of $x$ contains an open ball about $x$, and every open ball about $x$ contains (is!) a neighborhood of $x$.

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