Interior/Closure

**Deadstar** · May 4th 2009, 03:13 PM

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 [tex]X\setminus A[/tex] to get $X\setminus A$ .

**h2osprey** · May 5th 2009, 01:22 PM

Originally Posted by Deadstar

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 [tex]X\setminus A[/tex] 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.

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