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Let $(X,d)$ be a metric space and $A\subset X$. Prove that $\overline{A}=A\cup \partial A$ (that is, the closure of $A$ is the union of $A$ with its boundary).

Let $x\in \overline{A}$, then $x\in A$ or $x\not\in A$. Suppose $x\not\in A$, then $\forall\varepsilon >0, B_\varepsilon(x)\cap X\setminus A\ne \emptyset$. Since $\overline{A}$ is closed, there is a sequence $(x_k)_k\subset A$ converging to $x$. Since $(x_k)_k\subset A$, for every $x_k$ there exists $\varepsilon_{x_k}$ such that $B_{\varepsilon_{x_k}}(x_k)\subset A$. Thus $\forall \varepsilon >0$, $B_\varepsilon(x)\cap A\ne \emptyset$. Hence, $x\in \partial A$. Thus $\overline{A}\subset A\cup \partial A$.

For the other direction, suppose $x\in A\cup \partial A$, then $x\in A$ or $x\in \partial A$. If $x\in A$ then $x\in\overline{A}$. If $x\in \partial A$ then $\forall\varepsilon > 0$, $B_\varepsilon(x)\cap A\ne \emptyset \ne B_\varepsilon(x)\cap X\setminus A$. Since $B_\varepsilon(x)\cap A\ne \emptyset$, there are infinitely many points in $B_\varepsilon(x)$ of some sequence contained in $A$. Let $\varepsilon'<\varepsilon$, then there are infinitely many points in $B_\varepsilon'(x)$ of some sequence contained in $A$. Proceeding by induction and taking $\varepsilon$ smaller and smaller, we see that for every $\varepsilon > 0$ there are infinitely many terms in $B_\varepsilon(x)$ of some sequence contained in $A$. Inductively, this means that there is a sequence contained in $A$ which converges to $x$. Thus, $x\in\overline{A}$.

Please let me know if you find my proof correct. Maybe it's a bit over-complicated, but it would be interesting to know if it's OK.

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    $\begingroup$ What is the definition of $\partial$ that you are using? $\endgroup$ – Oiler Nov 2 '17 at 21:43
  • $\begingroup$ $\partial A$ is the boundary of $A$. $\endgroup$ – sequence Nov 2 '17 at 22:12
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    $\begingroup$ WLOG, A is open. $\endgroup$ – Jacob Wakem Aug 19 '18 at 18:07
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Note that $\partial A = \overline{A} \setminus A^\circ = \overline{A} \cap (A^\circ)^c$. Then $$A \cup \partial A = A \cup \left(\overline{A} \cap (A^\circ)^c\right) = (A \cup\overline{A}) \cap (A \cup (A^\circ)^c).$$ Since $A \subseteq \overline{A}$, $A \cup \overline{A} = \overline{A}$. Further, $\overline{A} \subseteq A \cup (A^\circ)^c$ so the result follows.

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  • $\begingroup$ Yes, this one is shorter. I didn't think of the set operation identity. $\endgroup$ – sequence Nov 2 '17 at 22:13
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Let $A^c$ be the complement of A. In any topological space we have $\partial A\subset \overline A$. Therefore $$\overline A= (\bar A\cap A)\cup (\bar A\cap A^c)=$$ $$=A\cup (\bar A\cap A^c)\subset$$ $$\subset A\cup (\bar A\cap \overline {A^c})=$$ $$=A\cup \delta A\subset$$ $$ \subset A\cup \bar A=\overline A.$$

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  • $\begingroup$ Thanks. Do you think my proof might be correct as well (albeit longer, of course)? My proof might be useful for those cases where we don't use the set theory identity. $\endgroup$ – sequence Nov 3 '17 at 1:24
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    $\begingroup$ I haven't read it over yet..... I dk why you would avoid the set theory. A property of all top'l spaces applies to all metric spaces and you can't say much about either of them without eventually using the simplest parts of set theory.. The general case often keeps you from becoming mired in details. And there are general results about top'l spaces that have specific uses for topics in the study of metric spaces. $\endgroup$ – DanielWainfleet Nov 3 '17 at 2:11
  • $\begingroup$ It's not that I avoid it, but I had completely forgotten about the identity. $\endgroup$ – sequence Nov 14 '17 at 15:57
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Write A= U {B(x,r(x))| x is in A} . cl A= U{ cl B(x,r)| x is in A} . Any boundary point is arbitrarily close to some x in A so it is within cl B(x,r). Thus every boundary point is in the closure.

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