Is the Limit Point of a Bounded Set Always a Supremum?

[lonewolf5999]

Hello,

I'm working through an analysis textbook on my own, and came across a true/false question I was hoping someone could help me with. The question is:

If A is a bounded set, then s = sup A is a limit point of A.

I think that the statement is false, as I came up with what I think is a counterexample. Let A = {1,2}, then clearly A is bounded since if a is an element of A, 1 <= a <= 2. Also, sup A = 2. But in an earlier exercise, I proved that finite sets have no limit points, so since A is finite, sup A = 2 is not a limit point of A even though A is bounded.

However, my book claims that this statement is true, and even gives a proof. I can't figure out what's wrong with my counterexample, so I'd really appreciate it if anyone could help me figure out what's wrong here. Thanks a lot!

 

[lavinia]

Hello,

I'm working through an analysis textbook on my own, and came across a true/false question I was hoping someone could help me with. The question is:

If A is a bounded set, then s = sup A is a limit point of A.

I think that the statement is false, as I came up with what I think is a counterexample. Let A = {1,2}, then clearly A is bounded since if a is an element of A, 1 <= a <= 2. Also, sup A = 2. But in an earlier exercise, I proved that finite sets have no limit points, so since A is finite, sup A = 2 is not a limit point of A even though A is bounded.

However, my book claims that this statement is true, and even gives a proof. I can't figure out what's wrong with my counterexample, so I'd really appreciate it if anyone could help me figure out what's wrong here. Thanks a lot!

Every element of a set is a limit point of the set. It is the limit of the constant Cauchy sequence. what you were probably thinking is that for a finite set there are no limit points outside of it.

 

[lonewolf5999]

Every element of a set is a limit point of the set. It is the limit of the constant Cauchy sequence. what you were probably thinking is that for a finite set there are no limit points outside of it.

I'd agree with you, except that my book also says that x is a limit point of a set A iff there exists a sequence (a_n) contained in A such that lim a_n = x, and a_n =/= x for all n. So this means that when looking for a sequence contained in A converging to x, we're not allowed to consider the constant sequence (a_n) = x, or a sequence which is eventually equal to x, or any other sequence which has x as an element, which is why a finite set has no limit points.

edit @ below: I hope that answers the question. Any thoughts on my counterexample?

 

[HallsofIvy]

That is not the standard definition of "limit point":
Wikipedia: "Let S be a subset of a topological space X. A point x in X is a limit point of S if every open set containing x contains at least one point of S different from x itself." (emphasis mine)
Using that definition, a finite set cannot have any limit points.
Lonewolf5999, does that text give a definition of limit point?

 

[lavinia]

I'd agree with you, except that my book also says that x is a limit point of a set A iff there exists a sequence (a_n) contained in A such that lim a_n = x, and a_n =/= x for all n. So this means that when looking for a sequence contained in A converging to x, we're not allowed to consider the constant sequence (a_n) = x, or a sequence which is eventually equal to x, or any other sequence which has x as an element, which is why a finite set has no limit points.

edit @ below: I hope that answers the question. Any thoughts on my counterexample?

This seems to be a matter of definition. Your counter example is correct if a limit point must be outside the set. Conceptually you are looking for all points that are limits of Cauchy sequences in the set. I guess you can divide these into those that are inside and those that are outside.

 

[lonewolf5999]

I'll just post the book's answer to this question and see if anyone has any thoughts on this as well.

"Let A be a bounded set, and let s = sup A. Then for any ε > 0, s - ε is not an upper bound, so we can find an element a of A such that a > s - ε. Hence a lies in the ε-neighborhood of s, and so the ε-neighborhood of s intersects A at a point other than s. Hence s is a limit point of A."

I think that this answer makes the assumption that s = sup A is not an element of A itself, but my counterexample doesn't, which is why sup A = 2 is not a limit point in my example. Is this all I need to resolve the difference between my answer and the book's answer?

 

[Dick]

I'll just post the book's answer to this question and see if anyone has any thoughts on this as well.

"Let A be a bounded set, and let s = sup A. Then for any ε > 0, s - ε is not an upper bound, so we can find an element a of A such that a > s - ε. Hence a lies in the ε-neighborhood of s, and so the ε-neighborhood of s intersects A at a point other than s. Hence s is a limit point of A."

I think that this answer makes the assumption that s = sup A is not an element of A itself, but my counterexample doesn't, which is why sup A = 2 is not a limit point in my example. Is this all I need to resolve the difference between my answer and the book's answer?

Yes. If you take the definition that a limit point is not an element of the set, then book's proof is not taking account of that clause in the definition.

 

[lonewolf5999]

Okay, I think that clears up the confusion I had. Thanks for helping me out.