Does Every Nonempty Proper Subset of R^n Have a Nonempty Boundary?

[redyelloworange]

Homework Statement

Prove that every nonempty proper subset of Rⁿ has a nonempty boundry.

The Attempt at a Solution

First of all, I let S be an nonempty subset of Rⁿ and S does not equal Rⁿ.

I tried to go about this in 2 different ways:

1) let x be in S and show that B(r,x) ∩ S ≠ ø and B(r,x) ∩ S^c≠ ø. I figured this wouldn't work with just one x in S. Or perhaps, I thought I should use induction on the number of elements in S?
2) Assume that bdS is empty and find a contradiction. However, I wasn't able to figure out a contradiction here. Unless, this implies that S equals Rⁿ, then that's a contradiction. But I'm not quite sure it implies that. I think that this is the proof you use to show that Rⁿ and the empty set are the only 2 that are both open and closed.

Thanks for your help! =)

Homework Statement

Homework Equations

The Attempt at a Solution

 

[NateTG]

What's the definition of boundry?

 

[redyelloworange]

bd(S) = {x in Rⁿ s.t. B(r,x)∩ S ≠ ø and B(r,x) ∩ Sc≠ ø for every r>0}

 

[Dick]

Take a point x in S and a point y in S^C and consider the line t*x+(1-t)*y for t in [0,1].

 

[river_rat]

Well you could assume the contrary and first prove that S must be closed and similarly that S must be open. Now which are the only sets in a connected space with that property?

 

[NateTG]

Well you could assume the contrary and first prove that S must be closed and similarly that S must be open. Now which are the only sets in a connected space with that property?

Redyelloworange is aware of that, he mentions that he thinks the goal is to prove that the n-dimensional reals are connected under the usual topolgy:

I think that this is the proof you use to show that Rn and the empty set are the only 2 that are both open and closed.