(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(R,d_{PO}) - i.e. with post-office metric

A=(0,1)

My question is how to find ext(A)

2. Relevant equations

ext(A)=int(A^{c})

d_{PO}(x,y)= |x|+|y| for x[tex]\neq[/tex]y and 0 for x=y

3. The attempt at a solution

As from the definition of an interior point, st.

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

A^{c}= (-[tex]\infty[/tex],0][tex]\cup[/tex][1,+[tex]\infty[/tex])

So we can create an open set st.;

Take [tex]\epsilon[/tex]=1/2 and a=0 and a=1, respectively;

B(0,1/2)={x [tex]\in[/tex] A^{c}: |x|< 1/2 for x [tex]\neq[/tex] 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 [tex]\neq[/tex] 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 [tex]\in[/tex] A^{c}: |x|< 1/2 for x [tex]\neq[/tex] 0 and 0< 1/2 for x=0}=(-1/2,1/2)

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

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# Homework Help: Exterior of a set

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