1. ## Interior/Closure

Is this correct?

Prove that $\displaystyle \overline{A} = X\setminus int(X\setminus A)$

Is it equivalent to prove that $\displaystyle X\setminus\overline{A} = int(X\setminus A)?$

If so then heres my proof...

Let $\displaystyle x \in X$\$\displaystyle \overline{A}$, then $\displaystyle B(x,r) \cap X$\$\displaystyle A = \emptyset$ $\displaystyle \forall r>0$
So then $\displaystyle \exists r > 0$ such that $\displaystyle B(x,r) \subseteq X$\A. Hence $\displaystyle x \in$ int(X\A).

Use $$X\setminus A$$ to get $\displaystyle X\setminus A$.

2. Originally Posted by Deadstar
Is this correct?

Prove that $\displaystyle \overline{A} = X\setminus int(X\setminus A)$

Is it equivalent to prove that $\displaystyle X\setminus\overline{A} = int(X\setminus A)?$

If so then heres my proof...

Let $\displaystyle x \in X$\$\displaystyle \overline{A}$, then $\displaystyle B(x,r) \cap X$\$\displaystyle A = \emptyset$ $\displaystyle \forall r>0$
So then $\displaystyle \exists r > 0$ such that $\displaystyle B(x,r) \subseteq X$\A. Hence $\displaystyle x \in$ int(X\A).

Use $$X\setminus A$$ to get $\displaystyle X\setminus A$.
Seems correct to me, except that it should be there exists instead of for all: there exists some s>0 such that the intersection is empty, that's the negation of for all r>0, the intersection is non-empty.