Show the $\mathbf{Int(A)}$ = $(\overline{A^{c}})^{c}$

Question

Show the $\ Int(A)$ = $(\overline{A^{c}})^{c}$

Definition we are using of closure: The closure of the set $A$ is defined to be the set $\overline{A}$ consisting of all limit points of $A$.

Attempt:

So my attempt was going to consist of letting $x\in Int(A)$ but I am not getting anywhere with that. I mean I know $Int(A)$ is open and taking the compliment of that is a close set, but I can't seem to see any light at the end of that attempt.

Answer 1

Recall the following definitions of closure and interior.

$\bar{B}$ is the smallest closed set containing B and $\operatorname{int}(B)$ is the largest open set contained in B. Perhaps you have covered the derivation of these definitions from the limit definitions already. Then

Note that since $\overline{A^c}$ is the smallest closed set containing $A^C$, its complement, ${\overline{A^c}}^c$ is the largest open set excluding $A^c$. Since $\operatorname{int}(A)$ is one such open set, we can see that $$\operatorname{int}(A) \subset {\overline{A^c}}^c$$

On the other hand, $\overline{A^c}$ is closed. Its complement ${\overline{A^c}}^c$ is open and is contained in $A$. Since $\operatorname{int}(A)$ is the largest such set (open, contained in $A$), we have that $$\operatorname{int}(A) \supset {\overline{A^c}}^c$$

Answer 2

One direction.

• $x \in Int(A)$

• There is a neighbourhood of $x$, call it $V$, such that $V$ is contained in $A$.

• If $x_n$ was a sequence of points in $A^c$ which converged to $x$, then this sequence would eventually have to enter $V$.

• This is a contradiction as $x_n \in A^c$ while $V$ is contained in $A$.

• Hence $x$ does not belong to the closure of $A^c$.

The other way:

• $x \notin Int(A)$.

• If $V$ is any neighbourhood of $x$ then $V$ intersects $A^c$ since it cannot be contained in $A$.

• So $x \in \overline{A^c}$, since every neighbourhood of $x$ is intersecting $A^c$.