Let $(X,d)$ a metric space and $A\subset X$. Prove that $$\overline{(A^c)}=\overset{\circ}{(A)}^c$$

i.e. the closure of complement, is the complement of the interior.

Proof. If $x\in \overline{(A^c)}$ then $x\in A^c$ or $x\in (A^c)'$. Since $\overset{\circ}{A} \subset A \Rightarrow A^c \subset \overset{\circ}{(A)}^c$ so, if $x\in A^c$ then $x\in \overset{\circ}{(A)}^c$.

Now if $x \not\in A^c$ but $x\in (A^c)'$ then $\forall r>0, B(x;r)-\{x\}\cap A^c \not= \emptyset.$ And now I've to deduce that $x\not\in \overset{\circ}{A}$ so $x\in \overset{\circ}{(A)}^c$, but I can't figure out how.

  • 1
    $\begingroup$ I posted an answer that solved the problem in an alternative way. But let me answer the question you actually asked based on the argument you gave. So you know that $B(x,r)$ intersects with $A^{c}$ for some value not equal to $x$. But this is true for every $r > 0$. So if that's the case, can we ever find an $r' > 0$ such that $B(x,r') \subseteq A$? No because every ball around $x$ intersects with $A^{c}$. But doesn't this mean $x \not \in A^{\circ}$? Since $A^{\circ}$ is those elements such that you can find a ball around them contained in $A$. So $x \in (A^{\circ})^{c}$, as you wanted $\endgroup$
    – layman
    Oct 12, 2014 at 22:12

4 Answers 4


If you use the topological definitions of closure and interior, it's very easy and natural: $$\overline A=\bigcap\,\{F\supseteq A\mid F\text{ is closed in }X\}$$ $$A^\circ=\bigcup\,\{G\subseteq A\mid G\text{ is open in }X\}$$ Now just use the rules for complements which turn the complement of a union into an intersection of the complements.

  • $\begingroup$ this is a good answer, and probably the only answer you should go to when there is no metric and only topology is concerned. $\endgroup$
    – Dinoman
    Sep 19, 2022 at 12:03

I think this is easier to do if you consider the closure $\overline{A}$ as the set of elements of $X$ such that any open ball around them intersects $A$, that is, $$\overline{A} = \{ x \in X \mid \forall \epsilon > 0, B(x,\epsilon) \cap A \neq \emptyset \}.$$ Similarly, one definition of interior is the set of elements of $X$ such that you can find an $\epsilon$-ball around them contained in $A$, that is,$$A^{\circ} = \{ x \in X \mid \exists \epsilon > 0 \text{ s.t. } B(x,\epsilon) \subseteq A \}.$$

Now, to prove the claim, first we will show $\overline{A^{c}} \subseteq (A^{\circ})^{c}:$

Let $x \in \overline{A^{c}}$. Then for every $\epsilon > 0$, $B(x, \epsilon) \cap A^{c} \neq \emptyset$. But this means any ball around $x$ will intersect with $A^{c}$, which means you can never find a ball around $x$ that is contained in $A$. That means $x \not \in A^{\circ}$. Thus, $x \not \in A^{\circ} \implies x \in (A^{\circ})^{c}$.

Now to show $(A^{\circ})^{c} \subseteq \overline{A^{c}}:$

Let $x \in (A^{\circ})^{c}$. Then $x \not \in A^{\circ}$. But this means there does not exist $\epsilon > 0$ such that $B(x,\epsilon) \subseteq A$. But this means for every $\epsilon > 0$, $B(x,\epsilon) \cap A^{c} \neq \emptyset$, and this is precisely what we need for $x \in \overline{A^{c}}$. So we are done.

  • $\begingroup$ +1 . Very elegant and smooth $\endgroup$ Aug 29, 2020 at 6:31
  • $\begingroup$ That proof could be shorter using an "iff way". $\endgroup$
    – LH8
    Nov 3, 2022 at 11:13

Here is a nice 'natural language proof', based on the facts that the interior of a set is the largest open set contained in it, and the closure of a set is the smallest closed set that contains it. By the way, this works for any topological space :

First, we note that $A^0$ is an open set contained in $A$, so $X\setminus A^0$ is a closed set that contains $X\setminus A$. So we know that $X\setminus A^0\supseteq \overline{X\setminus A}$. Now, assume that $X\setminus A^0$ is not the smallest closed set containing $X\setminus A$. Then, we would have some closed set $S$ such that $X\setminus A\subseteq S\subset X\setminus A^0$, which implies $A^0\subset X\setminus S\subseteq A$. Since $X\setminus S$ is open, this means that $A^0$ is not the smallest open set containing $A$, which is a contradiction.


Starting from the place where you are stuck:

Again $x \notin A^c$ so $B(x,r) \cap A^c \neq \phi \forall r$ So, $x\notin A^0$[From the definition of $A^0$].


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