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 $A$.