Suppose $A\subseteq X$. Prove that the boundary $\partial A$ of $A$ is closed in $X$.

My knowledge:

  • $A^{\circ}$ is the interior
  • $A^{\circ}\subseteq A \subseteq \overline{A}\subseteq X$

My proof was as follows:

To show $\partial A = \overline{A} \setminus A^{\circ}$ is closed, we have to show that the complement $( \partial A) ^C = X\setminus{}\partial A =X \setminus (\overline{A} \setminus A^{\circ})$ is open in $X$. This is the set $A^{\circ}\cup X \setminus(\overline{A})$

Then I claim that $A^{\circ}$ is open by definion ($a\in A^{\circ} \implies \exists \epsilon>0: B_\epsilon(a)\subseteq A$. As this is true for all $a$, by definition of open sets, $A^{\circ}$ is open.

My next claim is that $X \setminus \overline{A}$ is open. This is true because the complement is $\overline{A}$ is closed in $X$, hence $X \setminus \overline{A}$ is open in $X$.

My concluding claims are: We have a union of two open sets in $X$, By a proposition in my textbook, this set is open in $X$. Therefore the complement of that set is closed, which is we had to show.

What about this ?


From your definition, directly, $$ \partial A=\overline{A}\setminus \mathring{A}=\overline{A}\cap (X\setminus \mathring{A}) $$ is the intersection of two closed sets. Hence it is closed.

No need to prove that the complement is open, it just makes it longer and more complicated.

Also, keep in mind that a set $S$ is open in $X$ if and only if its complement $X\setminus S$ is closed in $X$. This should be pavlovian.

  • $\begingroup$ thank you for your reply.But is my proof correct ? $\endgroup$ – Applied mathematician Apr 5 '13 at 0:51
  • $\begingroup$ @JoyeuseSaintValentin Yes, it is correct. But aggain, it does not add anything interesting to consider the complement. Except for a little complication, i.e. determining this complement. So it is instructive for study purposes, but you should favor the direct way. $\endgroup$ – Julien Apr 5 '13 at 0:54
  • $\begingroup$ I agree. This was during an exam and I was not completely clear in my mind. Of course your proof is much 'cleaner' :) $\endgroup$ – Applied mathematician Apr 5 '13 at 0:58
  • $\begingroup$ @JoyeuseSaintValentin I've written crazy things during exams...And even at home without any pressure. $\endgroup$ – Julien Apr 5 '13 at 0:59
  • $\begingroup$ haha ok, good to hear that others struggle too sometimes $\endgroup$ – Applied mathematician Apr 5 '13 at 1:55

Your proof looks correct to me. The interior is indeed open by definition (and the closure is closed by definition). And the fact that a union of open sets is open is one of the axioms for a topology. You may wish to note that $\partial A$ is often defined as $\overline{A} \cap \overline{X - A}$. Since finite intersections of closed sets are closed, it follows that $\partial A$ is closed in $X$.


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