Prove alpha=sup(S) is equivalent to alpha belongs to S closure

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Given that alpha is an upper bound of a given set S of real numbers, prove that the following two conditions are equivalent:
a) We have alpha=sup(S)
b) We have alpha belongs to S closure

I'm trying to prove this using two steps.
Step one being: assume a is true, then prove b is true.
Step two being: assume b is true, then prove a is true.

Could anyone help me with step two?
Assuming alpha belongs to S closure...
 
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If I remember right, I think I gave you a useful condition for a point to be in the closure of a set. Do you see how you can use it here?
 
Of course, your "steps" are a correct way to prove equivalence of statements, from a logical point of view.
 
No I don't see how I can use it here in this problem.

How would I start my step two? I know I assume alpha belongs to S closure, but I am not sure where to go from there.
 
A point x is in the closure of a set A if any neighbourhood of x intersects A. Now, what is an important property of the supremum (involving "ε")?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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