# Open/closed subsets of metric space

1. Nov 25, 2011

### Ted123

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

3. 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?

2. Nov 25, 2011

### vela

Staff Emeritus
What definition of the boundary are you using?

3. Nov 25, 2011

### Ted123

$x\in\partial A$ if for all $r>0$, $B(x,r)$ intersects both $A$ and $A^c$.

Last edited: Nov 25, 2011
4. Nov 25, 2011

### vela

Staff Emeritus
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.

5. Nov 25, 2011

### Ted123

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?

6. Nov 25, 2011

### vela

Staff Emeritus
The answer should be pretty obvious if you consider the definition of a closed set.

7. Nov 25, 2011

### Ted123

Is every subset also closed?

If $x\in A^c$ then $B(x,1) = \{x\} \subseteq A^c$

8. Nov 25, 2011

### vela

Staff Emeritus
Yup. If every subset is open, Ac is open; therefore, A is closed.