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The following question is from Fred H. Croom's book "Principles of Topology"

Prove that the boundary of a subset $A$ of a metric space $X$ is always a closed set.

My attempt is as follows:

Theorem 3.10 states, the closure of a set is closed. Theorem 3.5 states, the intersection of any family of closed sets is closed. The boundary of a set is defined as $\bar{A}\cap\overline{(X\setminus A)}$; therefore, the boundary of $A$ is a closed set.

Granted, I did mention theorems from the book so it did most of the work, but is my proof okay? Or would I need to elaborate some more?


I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

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  • $\begingroup$ This is good. ${}{}{}$ $\endgroup$ – copper.hat Feb 18 '15 at 0:49
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For any $A \subset X \to X-\partial A$ an open set in $X$. To this end, let $x \in B = X-\partial A \to x \notin \partial A \to \exists r > 0: B(x,r) \subset \text{Int }A \text{ or } B(x,r) \subset X-\overline{A}$. But $X-\partial A = \text{Int } A \cup (X-\overline {A})$, so $B(x,r) \subset X-\partial A$, and $X-\partial A$ is open. Thus $\partial A$ is closed.

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Since BRIC-Fan already gave a solution, I'll give another one via sequences. Yes, I will kill a fly with a cannon.

Notice that: $$\begin{align} x \in \partial A &\iff \forall \, r > 0, \,B(x,r)\cap A \neq \varnothing \text{ and } B(x,r)\cap A^c \neq \varnothing \\ &\iff {\rm d}(x,A) = {\rm d}(x,A^c) = 0\end{align}$$

Also, we have that a set $E$ is closed if and only if $E = \overline{E}$. Furthermore, $x \in \overline{E}$ if and only if exists a sequence $(x_n)_{n \in \Bbb Z_{>0}}$ in $E$ such that $x_n \to x$. And if $X,Y$ are metric spaces, $f: X \to Y$ is continuous, and $x_n \to x$ in $X$, then $f(x_n) \to f(x)$ in $Y$. Now we're good to go.

Take a sequence $(x_n)_{n \in \Bbb Z_{> 0}}$ in $\partial A$, such that $x_n \to x$. If we prove that $x \in \partial A$, we're done. We have that ${\rm d}(x_n,A) = {\rm d}(x_n, A^c) = 0$ for all $n$. For any set $E \subset X$, the map $x \mapsto {\rm d}(x,E)$ is continuous. We take the limit: $$\lim {\rm d}(x_n,A) = \lim {\rm d}(x_n,A^c) = 0 \implies {\rm d}(x, A) = {\rm d}(x,A^c) = 0,$$ using continuity. So $x \in \partial A$, and hence $\partial A$ is closed.

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