Are There Only Two Clopen Subsets in ℝ: The Empty Set and ℝ Itself?

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Homework Help Overview

The discussion revolves around the properties of clopen subsets in the real numbers ℝ with respect to the absolute value metric. The original poster seeks to prove that the only clopen subsets are the empty set and ℝ itself, exploring the implications of a contradiction approach.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts a proof by contradiction, questioning the nature of a clopen subset A and its relationship with its complement A'. Participants discuss the implications of supremum and closed intervals, and whether certain points must belong to A or A'.

Discussion Status

Participants are actively engaging with the concepts of closed sets and limit points, with some clarifying misunderstandings about upper and lower bounds. There is a productive exchange regarding the properties of clopen sets and their complements, though no consensus has been reached on the proof's conclusion.

Contextual Notes

There is an ongoing exploration of definitions related to closed sets and limit points, as well as the implications of subsets being both open and closed. The discussion reflects a need for clarity on these foundational concepts without resolving the proof itself.

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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\inA}. Then I know b\inA or b\inA'.

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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SMA_01 said:
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?
 
SMA_01 said:
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?
 
SMA_01 said:
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?

Yes. Why is A' clopen too? (you just need that A' is closed by the way)
 
A' is clopen because A is both opened and closed. Thanks for your help!
 

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