# Interior/Closure

• May 4th 2009, 03:13 PM
Interior/Closure
Is this correct?

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

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

If so then heres my proof...

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

Use $$X\setminus A$$ to get $X\setminus A$.
• May 5th 2009, 01:22 PM
h2osprey
Quote:

Is this correct?

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

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

If so then heres my proof...

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

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

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.