Boundary of Subset A in Metric Space X: Proving Openness

[julypraise]

Homework Statement

Let $X$ be a metric space, and $A$ its nonempty proper subset. Then is $\partial A$ not open? If it is, how do I prove it?

Could you give me just some hints, not the whole solution?

Homework Equations

The Attempt at a Solution

I cannot even start..

 

[sunjin09]

Boundary is closed because it contains all of its accumulation points (why?). But boundary can be open too, since there are sets that are both open and closed (can you think of some example?)

 

[julypraise]

um... well, I've actullay solved this problem. Anin in my solution, I've proved that $\partial A$ is not open. Could u please check my solution?

$\partial A$ is closed:

$\overline{\partial A}=\overline{\overline{A}\cap\overline{X\backslash A}}\subseteq\overline{\overline{A}} \cap \overline{ \overline{X\backslash A}}=\partial A$

$\partial A$ is not open if it is not empty:

$\mbox{Int}\partial A=\mbox{Int}(\overline{A}\cap\overline{X\backslash A})=\mbox{Int}\overline{A}\cap\mbox{Int}\overline{X\backslash A}\subseteq\overline{A}\cap\overline{X\backslash A}=\emptyset$

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Ah! As I was writting this, I've realized I've made a silly mistake. I've concluded in the last line that the boundary of A is empty...

Anyway accodring to the part of the above observation, that is, the correct part of the above observation, the boundary of A is open when the closure of A and the closure of the complement of A are open.

I think a specific example can be either A=X or A = the empty set. But nontrivial example, um... cannot think of..

 

[HallsofIvy]

A slightly less trivial example is any set A with the discrete topology. That is, all subsets are both open and closed. It is the metric topology generated by the metric d(x,y)= a, for fixed number a, as long as $x\ne y$.

 

[julypraise]

A slightly less trivial example is any set A with the discrete topology. That is, all subsets are both open and closed. It is the metric topology generated by the metric d(x,y)= a, for fixed number a, as long as $x\ne y$.

Thanks HallsofIvy!

In this case, if I calculuate the boundary of A I get empty set because A is both open and closed.

I've found another example that if A=Q the set of rational numbers, then the boundary of A is R the set of real numbers.

It seems that if the boundary is open, which is a counterintutive fact, then it is either empty or the whole space...

Anyway I've found a web page that proves that if A is open or closed then the boundary of A is not open.

http://planetmath.org/encyclopedia/BoundaryOfAnOpenSetIsNowhereDense.html

(Actually this page proves that the boundary of A is nowhere dense.