# Exterior of a set

1. Jun 30, 2010

1. The problem statement, all variables and given/known data

(R,dPO) - i.e. with post-office metric

A=(0,1)

My question is how to find ext(A)

2. Relevant equations

ext(A)=int(Ac)

dPO(x,y)= |x|+|y| for x$$\neq$$y and 0 for x=y

3. The attempt at a solution

As from the definition of an interior point, st.

For every a$$\in$$A and $$\epsilon$$>0, there exists an open set B st. a$$\in$$B(a,$$\epsilon$$)$$\subset$$A then a is called an interior point of A.

Ac = (-$$\infty$$,0]$$\cup$$[1,+$$\infty$$)

So we can create an open set st.;

Take $$\epsilon$$=1/2 and a=0 and a=1, respectively;

B(0,1/2)={x $$\in$$ Ac : |x|< 1/2 for x $$\neq$$ 0 and 0< 1/2 for x=0}=(-1/2,1/2)

for this I am not sure wheter I should take x not equal to 0 or x=0. But for x not equal to 0, Ac obviously covers the open set, so I take x $$\neq$$ 0

and

B(1,1/2)={x in Ac : |x|+|1|< 1/2 for x not equal to 1 and 0< 1/2 for x=1} = {1} since |x|+|1|< 1/2 ---> |x|< -1/2 isn't correct. So 1 is included.

So, Ac covers each ball. That is, both 1 and 0 are the interior points of Ac and as a result;

int(Ac)= (-infty,0] (union) [1,+infty )

int(Ac)= (- infty ,0) (union) [1,+ infty )

I probably set some wrong logic here, could you help me why 0 is not included in int(Ac) ?

And for the open set;

B(0,1/2)={x $$\in$$ Ac : |x|< 1/2 for x $$\neq$$ 0 and 0< 1/2 for x=0}=(-1/2,1/2)

What if i take {0} instead of (-1/2,1/2)

Last edited: Jun 30, 2010
2. Jun 30, 2010

### Dick

You are right that B(0,1/2)=(-1/2,1/2). But A^C doesn't cover that ball. It also contains points from A. Can you show for any r>0 B(0,r) also contains points from A? That would say 0 is NOT in int(A^C) wouldn't it?

3. Jun 30, 2010

So you mean, since we cannot approximate zero by only the elements of Ac

for any r>0;

B(0,r)={x $$\in$$ Ac : |x|< r for x $$\neq$$ 0 and 0< r for x=0}=(-r,r)

Then for the interval (-r,r) around 0, there exist many elements from A.

Even though we can take r as small as we want, there exists elements from A. Thats why zero is not included, right?

4. Jun 30, 2010

### Dick

Right. And since B(1,1/2)={1} and 1 is not in A, 1 is included. Can you show any other point x of A^C has a ball around itself consisting only of x, just like 1?

5. Jun 30, 2010

B(|a|,|a|/2)={x $$\in$$ Ac : |x|+|a|< |a|/2 for x $$\neq$$ |a| and 0< |a|/2 for x=|a|}={|a|} since again |x|+|a|< |a|/2 is not correct.