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I am confused about definition of the boundary set. Definition: Given set $A$ and its complement $A^c$; Point $x$ belongs to the boundary of $A$ if every open ball centred at $x$ contains points of $A$ and $A^c$.

Now to my confusion: Taking as an example set $A\subset \mathbb{N}$ and its complement in $\mathbb{R}^1$. Set $A$ does not contain any limit points. Then for $\forall x\in A$ the only member of $A$ that is contained in all the balls centered at $x$, is $x$ itself. Now, $x\notin A^c$ but is a limit point in $A^c$, hence every ball centered at $x$ contains elements of $A^c$. Can I conclude that $x$ is in the boundary set of $A$, even though it is an isolated point in $A$? Also, say if the complement of $A$ is also in $\mathbb{N}$, will $x$ still be a boundary point of $A$?

Thanks a lot in advance! And I apologize for the clumsy language :-)


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Yes, being a boundary point means you can be an isolated point--you're reasoning is totally correct. As to your second question, I'm not quite sure what you mean. If $A\subseteq \mathbb{N}$ then $\mathbb{R}-A$ cannot be contained in $\mathbb{N}$ – Alex Youcis Nov 4 '11 at 5:32
up vote 2 down vote accepted

As pointed out by Alex Youcis, your reasoning is correct. What you can conclude also is that $\mathbb{N}$ is the boudary set of $\mathbb{R} - \mathbb{N}$. For your second question: No, if you consider $A \subset \mathbb{N}$ and $A^c = \mathbb{N} - A$, with the metric topology, then $\mathbb{N}$ is a discrete set so that no point of $A$ is the boundary of $A$

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Thanks Alex and Henrique – Leon Nov 6 '11 at 9:16

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