Find ext(A) for (R,dPO) with A=(0,1): Explanation and Solution

[caduceus]

Homework Statement

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

A=(0,1)

My question is how to find ext(A)

Homework Equations

ext(A)=int(A^c)

d_PO(x,y)= |x|+|y| for x\neqy and 0 for x=y

The Attempt at a Solution

As from the definition of an interior point, st.

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

A^c = (-\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 A^c : |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, A^c obviously covers the open set, so I take x \neq 0 and

B(1,1/2)={x in A^c : |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, A^c covers each ball. That is, both 1 and 0 are the interior points of A^c and as a result;

My answer is;

int(A^c)= (-infty,0] (union) [1,+infty )

But the given answer is;

int(A^c)= (- infty ,0) (union) [1,+ infty )

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

And for the open set;

B(0,1/2)={x \in A^c : |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)

 

[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?

 

[caduceus]

So you mean, since we cannot approximate zero by only the elements of A^c

for any r>0;

B(0,r)={x \in A^c : |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?

 

[Dick]

So you mean, since we cannot approximate zero by only the elements of A^c

for any r>0;

B(0,r)={x \in A^c : |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?

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?

 

[caduceus]

Actually it should work for and point in A^c except 0, for instance, for simplicity, we can take r=|a|/2 and we can take |a| instead of 1, then,

B(|a|,|a|/2)={x \in A^c : |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.

 

[Dick]

Actually it should work for and point in A^c except 0, for instance, for simplicity, we can take r=|a|/2 and we can take |a| instead of 1, then,

B(|a|,|a|/2)={x \in A^c : |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.

You've got the right idea. But you want B(a,|a|/2). B(|a|,|a|/2) isn't a ball around a, it's a ball around |a|.