If A is closed, then Bd(A) is a subset of A?

Question

I seem to be running in circles in this one. I know that A is a subset of Cl(A) so;

Bd(A) is a subset of A is a subset of Cl(A)

If A is closed then X-A is open.

Am in in the right ballpark? What do I do?

Dont put a link to a wikipedia page. I have read those and I have read my book hundreds of times.

Answer 1

A set is closed if it contains all of its boundary points. Since A is a closed set, it contains all of its boundary points. Therefore ∂A ⊆ A



You might be getting confused by the definition of a subset. If A ⊆ B, (if A is a subset of B) then A could equal B. This is different from A ⊂ B (A is a proper subset of B) which means A can't equal B.



Since A is closed, cl(A) = A

This means that cl(A) ⊆ A and A ⊆ cl(A)



Does that help any?

Answer 2

The definition I was provided was:



Bd(A) = Cl(A) \ Int(A)



So, by definition, Bd(A) is a subset of Cl(A), which equals A when A is closed. I've also read Wikipedia, and they list two other definitions. It's trivial for this definition as well:



Bd(A) = Cl(A) n Cl(X \ A)



Clearly Bd(A), being an intersection including Cl(A), must be a subset of Cl(A) (again which equals A when A is closed).



The third definition is less obvious. The boundary of A is defined as the set of points in the space having the property that every neighbourhood around the point contains both points in A and points not in A. But, based on this definition, we can find points in smaller and smaller neighbourhoods, and choosing points from these neighbourhoods will form a convergent sequence entirely in A. Thus the point in the boundary will be in Cl(A), and hence A, since A is closed.



Hope that helps!

Answer 3

The difference between an open set and a closed set is that the closed set includes its boundary and an open set does not. It's like the difference between < and less than or equal.My keyboard cant' make less than or equal.) An open set is akin to a limit.



Bd(A) is a subset of Cl(A). I think Bd(A) is a subset of A because A by itself is assumed to be closed.



If X is the universal set and A is closed, X-A is open because the boundary is part ofA

Answer 4

I would say that a closed set is a set that contains all its limit points. Since the Bd(S) contains all the limit points of S that are not in the interior of S that completes the proof.



I guess the problem is that there are about 5 different definitions of closed set and 5 different defns of boundary so you should probably tell us about what kind of definition in what kind of topological space.