Open/closed subsets of metric space

Ted123
Messages
428
Reaction score
0

Homework Statement



2mpx05s.jpg


The Attempt at a Solution



I've got through this question up to the last bit.

I've got B(0,1) = \{0\} and B(0,2) = \{y\in\mathbb{R} : -1<y<1 \} (i.e. the open interval (-1,1).)

How do I show that every subset of \mathbb{R} is open (A \subseteq X is open if it contains none if its boundary)

and then find which subsets are closed?
 
Physics news on Phys.org
What definition of the boundary are you using?
 
vela said:
What definition of the boundary are you using?

x\in\partial A if for all r>0, B(x,r) intersects both A and A^c.
 
Last edited:
OK, so that means if A is open, for every x in A, you can find some r>0 such that B(x,r) is contained in A.
 
vela said:
OK, so that means if A is open, for every x in A, you can find some r>0 such that B(x,r) is contained in A.

OK, so if A \subseteq \mathbb{R} and x\in A then B(x,1)=\{x\} \subseteq A so every subset A of \mathbb{R} is open.

How do I find which subsets are closed?
 
The answer should be pretty obvious if you consider the definition of a closed set.
 
vela said:
The answer should be pretty obvious if you consider the definition of a closed set.

Is every subset also closed?

If x\in A^c then B(x,1) = \{x\} \subseteq A^c
 
Yup. If every subset is open, Ac is open; therefore, A is closed.
 
Back
Top