## Clopen subsets of the reals?

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$\in$[x,y]|z$\in$A}. Then I know b$\in$A or b$\in$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.

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 Quote by SMA_01 I know that b is an upper bound for A implies b is a lower bound for A'.
That's not true is it? Take A=[a,b], then b is an upper bound of A. But $A^\prime=(-\infty,a)\cup (b,+\infty)$ and b is certainly not a lower bound of this.

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

 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?

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## Clopen subsets of the reals?

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

 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'?
 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?

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