Could someone help me get started with this one. I want to show that: bd(A)=Cl(A
ID: 2943147 • Letter: C
Question
Could someone help me get started with this one. I want to show that:
bd(A)=Cl(A)-Cl(X-A)
A is a subset of a topological space X, bd is the boundary of A
Explanation / Answer
>Let A be a subset of a metric space X. Prove that the boundary of A >is equivalent to A closure intersect with (X-A) closure > This is almost true by definition: a point x is in Bd(A) iff for all e>0 the ball B(x,e) intersects A and X-A. So this means in particular that (1) for all e>0 , B(x,e) intersects A But this says exactly that x is in cl(A). Also (2) for all e>0, B(x,e) intersects X-A And this says precisely that x is in cl(X-A). And if (1) and (2) hold, that is, x is in cl(A) AND in cl(X-A) (so in its intersection) than we have again that x is in Bd(A). So Bd(A) = cl(A) intersect cl(X-A). rate plz!