Clopen subsets of the reals?

SMA_01

Prove that the only subset of ℝ with the absolute value metric that are both open and closed are ℝ and ∅.

I know I'm supposed to prove by contradiction, but i'm having trouble:

Suppose there exists a clopen subset A of ℝ, where A≠ℝ, A≠∅. Let [x,y] be a closed interval in ℝ, where x is in A and y is in A' (complement of A). Now, let b=sup{z[itex]\in[/itex][x,y]|z[itex]\in[/itex]A}. Then I know b[itex]\in[/itex]A or b[itex]\in[/itex]A'.

I know that b is an upper bound for A implies b is a lower bound for A'. I'm just not sure how to arrive at a contradiction. I'm still not grasping the intuition behind it, can anyone explain intuitively what this means?

Thanks.

micromass

That's not true is it? Take A=[a,b], then b is an upper bound of A. But [itex]A^\prime=(-\infty,a)\cup (b,+\infty)[/itex] and b is certainly not a lower bound of this.

Anyway, by definition you know that b is the supremum of [itex][x,y]\cap A[/itex]. But the set [itex][x,y]\cap A[/itex] is closed (what is your definition of closed anyway?), what does that tel you about b?

SMA_01

Oh okay, I see my mistake.
Closed means a set contains its limit points. So if that intersection is closed, then b is in A?

micromass

OK, so b is an element of A. Can you make a similar argument to conclude that b is an element of A'?

SMA_01

Okay, b is an element of A because it is the intersection and A is closed. Why would it necessarily have to be in A'?

SMA_01

Unless, A' is clopen too right? So A' will have to contain all of its limit points as well, and b is a boundary point for A'...? Am I thinking about this correctly?

micromass

Yes. Why is A' clopen too? (you just need that A' is closed by the way)

SMA_01

A' is clopen because A is both opened and closed. Thanks for your help!