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Let $A\subset \mathbb{R}^n$ be a closed set. Prove that $\bar{A} = A$. ($\bar{A}$ is the closure of $A$, that is $\bar{A} = ∂A \cup A$, where $∂A$ is the boundary of $A$). Some propositions are just "too evident to prove", so I have no slightest idea how to start this proof. I've tried a proof by contradiction and tried to manipulate set operations, but to no avail so far. Would appreciate some hints.

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    $\begingroup$ How do you define $\bar{A}$? $\endgroup$ – Michael Albanese Jan 20 '16 at 20:58
  • $\begingroup$ The closure of $A$, that is $\bar{A} = ∂A \cup A$, where $∂A$ is the boundary. $\endgroup$ – sequence Jan 20 '16 at 21:00
  • $\begingroup$ @sequence how do you define a closed set? $\endgroup$ – JonSK Jan 20 '16 at 21:01
  • $\begingroup$ @JonSK: if $A^c$ is open then $A$ is closed. $\endgroup$ – sequence Jan 20 '16 at 21:01
  • $\begingroup$ How do you define $\partial A$? $\endgroup$ – Sten Jan 20 '16 at 21:02
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It is clear that $A\subseteq\overline{A}$.

Suppose that $A^c$ is open. Take $x\in\partial A\cup A$. If $x\in A$ we are done. If $x\in\partial A$ then it must be $x\in A$ too, otherwise $x\in A^c$ and as $A^c$ is open then $A^c\cap A\neq\emptyset$ (because $x\in\partial A$) which is absurd. Therefore $\overline{A}\subseteq A$.

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  • $\begingroup$ I don't get the argument that if $x\in A^c$ then $A^c \cap A \ne \emptyset$. If $x\in ∂A$ then how does this imply that $A^c \cap A \ne \emptyset$? $\endgroup$ – sequence Jan 20 '16 at 21:21
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    $\begingroup$ Because $A^c$ is open, if $x\in A^c$ it contains an open ball $B_{\epsilon}(x)$. Then if $x\in\partial A$ we have $\emptyset\neq B_{\epsilon}(x)\cap A\subseteq A^c\cap A$. $\endgroup$ – JonSK Jan 20 '16 at 21:51
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It suffices to show that $\partial A \subseteq A$.

Let $x \in \partial A$. By definition, every neighbourhood of $x$ contains points from both $A$ and $A^c$.

Assume for a contradiction that $x \not \in A$. Then $x \in A^c$.

Since $A$ is closed, $A^c$ is open, and therefore $x$ has a neighbourhood that is contained in $A^c$. In other words, $x$ has a neighbourhood which does not contain any points from $A$. This contradicts the definition of the boundary. The result follows.

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