Using int A = X\clA to Deduce clA = X\intA

[mona88]

X is a metric space and A \subseteq X
Use int A = \cup{0\subseteqX : O open and O \subseteqA} to deduce that clA= \cap{C\subseteqX : C closed and A \subseteq C}.


i have proved that int(X\A) = X\(clA) and cl(X\A) = X\(intA)


I tried to prove it starting with x\in the intersection of C, where C is closed
if and only if x \notin X\C, where X\C is open
Then B(x, r) \notin X\C, where B(x,r) is the open ball, centre x, radius r,
so, x \notin int(X\C), and hence x \notin X\clC
therefore x \incl C. And hence,must also be in the closure of A.

 

[Gib Z]

X is a metric space and A \subseteq X
Use int A = \cup{0\subseteqX : O open and O \subseteqA} to deduce that clA= \cap{C\subseteqX : C closed and A \subseteq C}.


i have proved that int(X\A) = X\(clA)

So you can say clA = X\int(X\A) . Use that and De Morgan's Laws to prove the result.