# Proof of equality of diameter of a set and its closure

• chipotleaway
In summary, one method of proving diam(cl(A)) ≤ diam(A) is by letting x,y be points in cl(A) and showing that for any radius r>0, there exist balls B(x,r) and B(y,r) that intersect with A. However, there is a potential issue with isolated points, so it is better to let x,y be in the set of limit points of A. This leads to the conclusion that sup{d(x,y): x,y in the set of limit points of A} ≤ 2r+diam(A). The only problem remaining is with points in cl(A) that are not in A.
chipotleaway
In showing diam(cl(A)) ≤ diam(A), (cl(A)=closure of A) one method of proof* involves letting x,y be points in cl(A) and saying that for any radius r>0, balls B(x,r) and B(y,r) exist such that the balls intersect with A.

But if x,y is in cl(A), isn't there the possibility that x,y are isolated points? Shouldn't we let x,y be in the set of limit points of A?

But doing that, I get (a is in the intersection of B(x,r) and A, b is in the intersection of B(y,r) and A).

d(x,y)≤d(x,a)+d(a,b)+d(b,y)<2r+d(a,b)
Then taking the supremum of both sides
∴ sup{d(x,y): x,y in the set of limit points of A} ≤ 2r+sup{d(a,b): a,b in A}
∴ sup{d(x,y): x,y in the set of limit points of A} ≤ 2r+diam(A)

The problem being the left hand side doesn't become the definition of diam(cl(A))

Sure, ##x## and ##y## can be isolated points. But then you know that ##x,y\in A## for sure. So ##B(x,r)## and ##B(y,r)## intersect ##A##. Hence the proof continues like usual.

1 person
Ah, thanks micromass. If x and y were assumed to be isolated points it would just follow that diam(A)=diam(cl(A)) cause they're in A right?

chipotleaway said:
Ah, thanks micromass. If x and y were assumed to be isolated points it would just follow that diam(A)=diam(cl(A)) cause they're in A right?

Yeah, clearly the only problem is with points which are in ##\mathrm{cl}(A)## but not in ##A##.

Firstly, it is important to clarify that the equality of diameter of a set and its closure is a well-known and established fact in mathematics. It is not just a possibility, but a proven result. Therefore, we can trust the validity of the proof provided in the link.

To address the concern about x and y potentially being isolated points, it is true that in general, the closure of a set may contain isolated points. However, in this specific case, we are considering the closure of a set A, which means that all points in the closure are either in A or are limit points of A. Therefore, we can safely assume that x and y are in the set of limit points of A.

Furthermore, the proof provided in the link makes use of the fact that for any point in the closure, there exists a ball of a certain radius that intersects with the set A. This is a property of the closure, and it ensures that the points x and y are close enough to A for the proof to hold.

In addition, the proof considers the supremum of the distances between all possible pairs of points in the set of limit points of A. This is a valid approach, as the supremum of a set is the smallest upper bound, meaning that it takes into account all possible distances between limit points of A.

Therefore, the left-hand side does indeed represent the definition of diam(cl(A)), and the proof is valid.

## 1. What is the concept of "equality of diameter of a set and its closure"?

The equality of diameter of a set and its closure is a mathematical concept that states that the diameter of a set and the diameter of its closure are equal. This means that the distance between any two points in a set and the distance between those same two points in the closure of the set are the same.

## 2. How is the proof of equality of diameter of a set and its closure useful?

The proof of equality of diameter of a set and its closure is useful in many areas of mathematics, particularly in topology and analysis. It allows us to make conclusions about the behavior of sets and their limits, and to prove theorems and properties of these sets.

## 3. Can you provide an example of a set and its closure that demonstrate the equality of diameter?

Yes, let's consider the set of all rational numbers between 0 and 1, denoted by Q. The closure of this set, denoted by Q̅, would include all irrational numbers between 0 and 1 as well. Since the rational numbers are dense in the real numbers, we can see that the diameter of Q and Q̅ are equal, as the distance between any two points in Q is the same as the distance between those same two points in Q̅.

## 4. How is the proof of equality of diameter of a set and its closure different from the proof of density of a set?

The proof of density of a set is a separate concept that states that a set is dense in a space if every point in the space can be approximated by a point in the set. This is different from the proof of equality of diameter, which specifically focuses on the distance between points in a set and its closure. While both concepts are important in topology and analysis, they are distinct from each other.

## 5. Are there any limitations or exceptions to the proof of equality of diameter of a set and its closure?

Yes, there are some cases where this proof may not hold. For example, if a set is not closed, then its closure may have a different diameter. Additionally, in some non-metric spaces, the notion of diameter may not be well-defined, making this proof irrelevant. However, in most standard mathematical spaces, the proof of equality of diameter is a valid concept.

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