Outer area of S = outer area of S closure, analysis

[missavvy]

Homework Statement

Show that the outer area of S = outer area of the closure of S

Homework Equations

The Attempt at a Solution

I don't really understand 100% the difference between the set S and the closure of S.
I know the closure is S $$\cup$$ $$\partial$$S (the boundary of S), but then what does that mean S is? Since S is not the same as the interior of S or the closure..

Anyways,
Let A denote the outer area.

I'm starting by showing that A(S) $$\leq$$ A(Cls).

Let a partition of a rectangle which contains R, P= {x_i, y_j} Then
S = upper sum, 1 = characteristic function

S_p(1_s) = $$\sum$$ M_ij A(R_ij)

Where M_ij = sup{1_s(x,y); (x,y) belong to Rij}

I'm trying to figure out how the rectangles are defined in S, is it just R_ij $$\cap$$S?
Then
S_p(1_s) $$\leq$$ $$\sum$$ M_ij A(R_ij) = S_p1_[Cls(S)]
R_ij $$\cap$$ S & R_ij $$\cap$$ $$\partial$$S
Anyways once I show that A(S)$$\leq$$ A(Cls), I think the other inequality will be harder.. any tips?

If possible, could someone maybe describe it as a picture? I find I understand this stuff much better visually..

Thanks :)

 

[Robert1986]

Let's think in R, for a minute. And let's think about the open interval (0,1) in particular. Now, think about placing a divider to the left of 0 and one to the right of 0. Now slide the two dividers together as far as they can go without actually touching either a point of (0,1) or a boundary point of (0,1). The stuff between the dividers is the closure of (0,1). In other words, it is kind of like putting a fence on the boundary points of a set and then saying everything on the inside of the fence, and the fence it self is the closure of that set.

So, open sets do not contain their boundaries, that is the difference.

Can you show that the boundary of a set S takes up no "area" (whatever "area" is)?

 

[missavvy]

Yes, if the boundary has zero content. So if were to show that the area of the boundary, which is a part of the closure, has no area then they would be equal.

 

[Robert1986]

Yes, if the boundary has zero content. So if were to show that the area of the boundary, which is a part of the closure, has no area then they would be equal.

Correct, since the closure of a set consists of the boundary points of the set and the sets interior points.