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

In summary, the conversation is about proving the equivalence of two conditions regarding an upper bound alpha of a set S of real numbers. The first condition is that alpha is equal to the supremum of S, and the second condition is that alpha belongs to the closure of S. The person asking for help is trying to prove this using two steps, and is asking for assistance with the second step. The responder suggests using the definition of a point being in the closure of a set, and hints at using the property of the supremum involving "ε" to continue the proof.
  • #1
splash_lover
11
0
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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  • #2
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?
 
  • #3
Of course, your "steps" are a correct way to prove equivalence of statements, from a logical point of view.
 
  • #4
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.
 
  • #5
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")?
 

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

1. What is the definition of alpha=sup(S)?

The definition of alpha=sup(S) is the supremum of a set S, which is the least upper bound of all elements in S. This means that alpha is the smallest number that is greater than or equal to every element in S.

2. What is the definition of alpha belongs to S closure?

The definition of alpha belongs to S closure is that alpha is an element of the closure of set S, which is the smallest closed set that contains all elements of S. This means that alpha is either an element of S or a limit point of S.

3. How is proving alpha=sup(S) equivalent to alpha belongs to S closure?

Proving alpha=sup(S) is equivalent to alpha belongs to S closure by showing that alpha satisfies both definitions. In other words, alpha is the smallest number that is greater than or equal to every element in S, and it is either an element of S or a limit point of S. This proves that alpha belongs to the closure of set S.

4. What does it mean for alpha to be a limit point of S?

A limit point of S is a point that is arbitrarily close to the elements of S. This means that for any given distance, there exists an element of S within that distance from alpha. It does not necessarily mean that alpha is an element of S, but it is a point that is "approaching" S.

5. Can alpha=sup(S) be proven without showing that alpha is a limit point of S?

Yes, alpha=sup(S) can be proven without showing that alpha is a limit point of S. This is because the definition of supremum only requires that alpha is the smallest number that is greater than or equal to every element in S. However, showing that alpha is a limit point of S can help to strengthen the proof and provide a deeper understanding of the relationship between alpha and S.

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