Proof of equality of diameter of a set and its closure

[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))


*https://www.physicsforums.com/showthread.php?t=416201 (post #6)

 

[micromass]

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.

 

[chipotleaway]

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?

 

[micromass]

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 $A$.

 

[blue_raver22]

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.