Explanation of Limit Points

[G-X]

I have a few questions of about limit points:

Definition: The closure of a set A is [Math] \bar A = A ∪ A′[/Math], where [Math]A′[/Math] is the set of all limit points of A.

1) A′ - How can this be the set of all limit points of A - if A contains limit points?

2) Also, A closed set is one which contains all its limit points - How is this possible if A being open and compliment contains limit points

 

[pka]

Definition: The closure of a set A is [Math] \bar A = A ∪ A′[/Math], where [Math]A′[/Math] is the set of all limit points of A.
1) A′ - How can this be the set of all limit points of A - if A contains limit points?
2) Also, A closed set is one which contains all its limit points - How is this possible if A being open and compliment contains limit points FALSE

Consider this set in \(\Re^1\), \(G=(1,\infty)\) Now \(G\) is an open set. Every point of \(\bf G\) is a limit point of \(\bf G\).
So \(G\subset G'\) But \(\overline{~G~}=[1,\infty)=G\cup \{1\}\) is the closure of \(G\).
The complement (note the spelling) of \(G\) is \((-\infty,1]\).

 

[Steven G]

1) Why can't both A and A' both contain limit points? A' just 'adds' the limits points needed 'to close A'.

 

[G-X]

oh... I read it "is the set of all limit points of A" not additional limit points... simply all the limit points, that exist, that are in A... So the wording has me confused... and I cannot understand the meaning...

Should I read it - such that it means additional limit points of A - the limit points that are missing and thus could read - all the limit points of A necessary to close A? From your meaning and I assume you take it from the closure definition, the fewest amount of limit points necessary to close A - such that you get both interior and exterior...

 

[pka]

1) Why can't both A and A' both contain limit points? A' just 'adds' the limits points needed 'to close A'.

The quick answer is that \(A~\&~A^{\prime}\) can both contain limits points of \(A\) but \(A^{\prime}\) contains all the limits.
Moreover \(A^{\prime}\) is a closed set. For some set \(H\) if \(H^{\prime}\subset H\) then the set \(H\) is closed.

 

[G-X]

If A' contains all the limits how can H' ⊂ H as it would have more elements? Thus rather it is the other way around H ⊂ H'. As H' has all the limits and more.

How does A' contain all the limits when it is just to serve the purpose of closing A - the boundary points.

The closure of A is suppose to contain the exterior and interior points and all the limit points

 

[Dr.Peterson]

oh... I read it "is the set of all limit points of A" not additional limit points... simply all the limit points, that exist, that are in A... So the wording has me confused... and I cannot understand the meaning...

Should I read it - such that it means additional limit points of A - the limit points that are missing and thus could read - all the limit points of A necessary to close A? From your meaning and I assume you take it from the closure definition, the fewest amount of limit points necessary to close A - such that you get both interior and exterior...

No, in the definition you are using, A' is the set of all limit points of A. There is no need to omit limit points that are in A; the union of the two sets doesn't get confused if a point is in both A and A'! It will still be counted only once.

Suppose [MATH]A = (0,1)\cup\{2\}[/MATH]. Then [MATH]A' = [0,1][/MATH]; it is [MATH]A[/MATH] together with [MATH]\{0\}[/MATH] and [MATH]\{1\}[/MATH], but excluding [MATH]\{2\}[/MATH]. Their union is [MATH][0,1]\cup\{2\}[/MATH].

 

[G-X]

Why devise this thing called A' with these extra elements that are already in A? When you can just say A' is the boundary elements or limit points that suffice to close A and make [Math]\bar A[/Math]

 

[Dr.Peterson]

Because it's far simpler to just say the closure consists of A together with its limit points, without making you pick out those that are not already in A. That's part of the value of the union concept.

Note that many definitions of closure make no mention of this A', which has no importance of its own.

Mathematicians avoid extra words.

 

[G-X]

I am not sure how that is relevant. Denoting something as just boundary points shouldn't make you need to pick anything out? Does it somehow make it easier to solve a problem?

 

[pka]

If A' contains all the limits how can H' ⊂ H as it would have more elements? Thus rather it is the other way around H ⊂ H'. As H' has all the limits and more. How does A' contain all the limits when it is just to serve the purpose of closing A - the boundary points. The closure of A is suppose to contain the exterior and interior points and all the limit points

TO: G-X your real problem is the lack of ability to read and/or follow mathematical language. The statement is if \(H^{\prime}\subset H\) then the set is closed. In other words: closed sets contain all of their limit points.
Consider these open intervals, \(\forall n\in\mathbb{Z}^+ \mathcal{O}_n=\left(\dfrac{1}{n+1},\dfrac{1}{n}\right)\).
Is it clear to you that \((\forall n)~\mathcal{O}_n\subset [0,1]~?\) Hence \( \mathcal{J}=\bigcup\limits_{n = 1}^\infty {{O_n}}\subseteq [0,1]. \)
BUT do you see that \(\mathcal{J}\) is an open set? If so explain; if not tell us why not.
Do you see the for \((\forall n\in\mathbb{Z}^+)\left[\dfrac{1}{n}\notin\mathcal{J}\right]\) but \(\dfrac{1}{n}\in\mathcal{J}^{\prime}\).

 

[Steven G]

A may have no limit points, some limit points or all its limit points. A', by definition, must have all limit points of A. A' does not only have the limit points that A does not have. Sorry for confusing you.

 

[Dr.Peterson]

I am not sure how that is relevant. Denoting something as just boundary points shouldn't make you need to pick anything out? Does it somehow make it easier to solve a problem?

It makes it a little easier to state their definition. As I said, this is not the only way to define closure. When you write your book, you may choose to do differently, and it will not (necessarily) be wrong.

But when you are reading someone's book, you use their definitions. If it says A' is the set of all limit points of A, then that's what it is. That's how definitions work. Nothing they say is wrong; it's just different from what you're expecting. That suggests that you may have something to learn from this author about how to think about these ideas.

What I'm more concerned about is that you seem to think the situation stated in the definition is impossible. That is what you need to think about. Go back and see if you can answer your questions in post #1. I'd recommend using an example, as I did in post #7.