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$$\subseteq$$X : O open and O $$\subseteq$$A} to deduce that clA= $$\cap$${C$$\subseteq$$X : 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 $$\in$$cl 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$$\subseteq$$X : O open and O $$\subseteq$$A} to deduce that clA= $$\cap$${C$$\subseteq$$X : 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.