Closure of the complement, iteration, set identity

Question

Let $A$ be a closed subset of a topological space $X$ and $A'=\overline{X\setminus A}$. Then holds $A'''=A'$

I want to proof this equality. I might have $\subseteq$.

First of all $A'''=\overline{X\setminus (\overline{X\setminus (\overline{X\setminus A})})}$.

I want to show $A'''=A'$

Let $x\in A'''$. Then $x\in\bigcap_{U\supseteq X\setminus (\overline{X\setminus (\overline{X\setminus A})})\\ U\, \text{closed}} U$

And similar:

$U\supseteq X\setminus \bigcap_{V\supseteq X\setminus (\overline{X\setminus A)}} V$

$V\supseteq X\setminus \bigcap_{W\supseteq X\setminus A} W$

Now suppose $x\notin \overline{X\setminus A}$.

[Basically it goes like this now. I spare you the whole argument here, it should be clear how it is done at this point, I hope.]

Then $x\notin\bigcap W\Rightarrow x\in V\Rightarrow x\notin U\Rightarrow x\notin\bigcap U$. Contradiction.

If this is wrong, let me know and I provide a rigorouse proof, if this is not understandable.

Now I want to proof $A'''\supseteq A'$.

But here I am kinda stuck... Espacially I do not see where to involve that $A$ is closed.

Hints are appreciated. Thanks in advance.

Answer 1

Let $C(s)=\overline s$ and $I(s)=int(s).$

Theorem: $CICI(s)=CI(s).$

Proof: (i). $CICI(s)=C(I(CI(s))\subset C(CI(s)) =CI(s).$

(ii). $CICI(s)=CI(C(I(s))\supset CI(I(s))=CI(s).$

Let $s^-=\overline s$ and $s^c=X$ \ $s.$ With the superscripts acting from left to right we have $CI(s)=s^{c-c-}$ and $s'=s^{c-}=C(s^c).$

So $A'''=A^{c-c-c-}.$ Now since $A=A^-$ we have $A^c=I(A^c).$ Therefore $$ A' =C(A^c)=$$ $$=CI(A^c)= CICI(A^c)=$$ $$=CIC(I(A^c))=CIC(A^c)=$$ $$=CI(A^{c-})=$$ $$= (A^{c-})^{c-c-}=A^{c-c-c-}=A'''.$$

Sleight of hand: Change $A^c$ to $I(A^c)$ from the 1st to 2nd line in order to apply the Theorem , and then change $I(A^c)$ back to $A^c$ in the third line.

Answer 2

int A, interior of A;  cl A, closure of A;  S topological space. 
 
int cl A = int cl int A.  Algebraic (set algebra) proof: 
. . int cl A = int int cl A subset int cl int cl A  
. . subset int cl cl A = int cl A 
 
cl S\A = S - int A  A point by point proof can be done with a 
series of a half dozen equivalent statements:  
x in cl S\A iff ... iff x in S - int A. 
 
Algebraic proof of A' = A"': 
A' = cl S\A = S - int A 
A" = cl int A 
A"' = cl (S - cl int A) = S - int cl int A 
. . = S - int cl int cl A 
. . = S - int cl A = S - int A = A'. 
 
Where is the fact that A is closed, used?