Prove $C(A)=A\cup\partial A$.

I need to prove inclusion on both sides. Here is my attempt:

($\rightarrow$) $C(A)\subseteq A\cup\partial A$

Let $x\in C(A)$. Suppose $x\not\in A\rightarrow A$ is open. Since $x\in C(A)$, for $\epsilon>0,$ $B_\epsilon (x) \cap A\neq \emptyset$. Also since $A^c$ is closed, for $\epsilon>0,$ $B_\epsilon (x)/{x} \cap A^c\neq \emptyset$ (dubious?) Hence $x\in\partial A$.

($\leftarrow$) $A\cup\partial A\subseteq C(A)$
If $x_0\in A$ then by definition of closure it is in $C(A)$. So let $x_o\in\partial A/A$. Suppose $x_o\in C(A)^c$, which is open. Then $\exists\epsilon>0$ s.t. $B_\epsilon (x) \cap C(A)^c \neq \emptyset$ but by definition of boundary then $B_\epsilon (x) \cap C(A)\neq \emptyset$... not sure how to carry this over to $x\in\ C(A)$.

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    $\begingroup$ I don't see how "$x\notin A\to A$ is open" makes any sense - $A$ is either open or it isn't. $\endgroup$ – Zev Chonoles Feb 3 '12 at 20:40
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    $\begingroup$ @Emir: No. $1$ is not in $[0,1)$, but $[0,1)$ is not open in $\mathbb{R}$. $\endgroup$ – Brian M. Scott Feb 3 '12 at 20:47
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    $\begingroup$ $x$ is in the closure of $A$ and $x\notin A$ tells you $A$ is not closed. But, sets are not doors... $\endgroup$ – David Mitra Feb 3 '12 at 21:17
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    $\begingroup$ I’m inclined not to close, since a full answer should address Emir’s specific argument. $\endgroup$ – Brian M. Scott Feb 3 '12 at 21:26
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    $\begingroup$ @Emir How do you define the boundary of $A$? $\endgroup$ – user38268 Feb 4 '12 at 3:21

Using your notation $C(A)$ for the closure of $A$, and working in a metric space $X$:

$A \subset C(A)$ because if $x$ is in $A$, then any $B_{\epsilon}(x)$ for $\epsilon > 0$ intersects $A$ (it always contains $x$ from $A$..), so $A \subset C(A)$.

$\partial A \subset C(A)$: if $x \in \partial A$, then for any $\epsilon > 0$, $B_{\epsilon}(x)$ intersects $A$ (and also intersects $X \setminus A$) by the definition of the boundary of $A$, so $x \in C(A)$.

As both $A$ and $\partial A$ are a subset of $C(A)$, so is their union. This is part 1.

Assume $x \in C(A)$; we want to show $x \in A \cup \partial A$. If $x$ is in $A$, we are done, so assume $x \notin A$. Now, for every $\epsilon > 0$, $B_{\epsilon}(x)$ intersects $A$ (as $x \in C(A)$) and it also intersects $X \setminus A$, because it contains $x$ and we assumed $x \notin A$. This shows that $x \in \partial A$ (when $x \notin A$), and together this means $C(A) \subset A \cup \partial A$, concluding the proof.


$\boxed{C(A) \subset A \cup \partial A}$

As others have said in the comments, you can't conclude that $A$ is open simply because there is an element $C(A) - A$. It is true that $C(A) \subset A$ if and only if $A$ is closed, but there are lots of sets which are neither closed nor open. For a concrete example, look at $A = [0, 1)$ in the real line. The point $1$ is in $C(A) - A$, and yet $A$ is not open (Why?).

But you can still use some of your ideas. If $x \in C(A) - A$ then, as you say, for each $\epsilon > 0$ the ball $B_\epsilon(x)$ intersects $A$, and note that $x \in A^c$. Perhaps the definition of a limit point, which involves looking at $B_\epsilon(x) - \{x\}$, is entering your thoughts, but we don't need to use that here.

$\boxed{A \cup \partial A \subset C(A)}$

I don't see why we're taking the complement of $C(A)$, having fixed $x_0 \in \partial A - A$. The idea of taking complements does suggest proving the equivalent (it's the contrapositive) statement: $C(A)^c \subset (A \cup \partial A)^c$. For this, if $x_0$ is contained in the open set $C(A)^c$ then there is an $\epsilon > 0$ such that $B_\epsilon(x_0) \subset C(A)^c \subset A^c$, and hence $B_\epsilon(x_0) \cap A = \varnothing$. Can you use this to conclude that $x_0$ is neither an element of $A$ nor of the boundary?

  • $\begingroup$ I did not see Brian's "closed but not open" example until after submitting. Eerie! Although I guess it's the obvious thing to write down. $\endgroup$ – Dylan Moreland Feb 3 '12 at 23:13

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