
#1
Nov2712, 01:17 AM

P: 218

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. 



#2
Nov2712, 01:23 AM

Mentor
P: 16,542

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? 



#3
Nov2712, 06:16 AM

P: 218

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? 



#4
Nov2712, 08:34 AM

Mentor
P: 16,542

Clopen subsets of the reals? 



#5
Nov2712, 09:15 AM

P: 218

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'?




#6
Nov2712, 09:17 AM

P: 218

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?




#7
Nov2712, 10:59 AM

Mentor
P: 16,542





#8
Nov2712, 09:42 PM

P: 218

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



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