How does the diameter of a closure relate to isolated points in Rudin's book?

[Bachelier]

In Rudin we read $diam \ \bar{S} = diam \ S$.

And the $2ε$ trick is very clear. However I see how would this would work for an accumulation point of $S$ but what about an Isolated point of $S$ that is miles away from the set.

 

[micromass]

Could you post the proof that Rudin gives??

 

[Bachelier]

Could you post the proof that Rudin gives??

$Diam \ S ≤ Diam \ \bar{S}$ is trivial

Now consider 2 points $p, \ q \in \ \bar{S}$. Then there exists $p', \ q' \in \ {S}$ for which:

$d(p,p')< ε \ and \ d(q,q') < ε, \ for \ a \ given \ ε > 0$ (This is the definition of $\bar{S}$)

$So\ now: \ d(p,q)≤d(p,p')+d(p',q')+d(q,q')$

$=> d(p,q)<2ε+d(p',q')$

$=> d(p,q)<2ε +Diam \ S$

hence $Diam \ \bar{S} ≤ 2ε + Diam \ S$

since ε is arbitrary, the result is proven.

 

[HallsofIvy]

What problem do you have with isolated points? The diameter of set A, as well as the diameter of its closure, depends upon the entire set, not individual points.

If, for example, $A= (0, 1)\cup {2}$ then, since we have points arbitrarily close to 0 in the set, the diameter of A is 2- 0= 2. The closure of A is, of course, $[0, 1]\cup {2}$ which still has diameter 2. Another example is $A= (0, 1)\cup {2}\cup (3, 4)$ whicy has diameter 4- 0= 4. It's closure is $[0, 1]\cup {2}\cup [3, 4]$ which still has diameter 4.

 

[micromass]

$Diam \ S ≤ Diam \ \bar{S}$ is trivial

Now consider 2 points $p, \ q \in \ \bar{S}$. Then there exists $p', \ q' \in \ {S}$ for which:

$d(p,p')< ε \ and \ d(q,q') < ε, \ for \ a \ given \ ε > 0$ (This is the definition of $\bar{S}$)

$So\ now: \ d(p,q)≤d(p,p')+d(p',q')+d(q,q')$

$=> d(p,q)<2ε+d(p',q')$

$=> d(p,q)<2ε +Diam \ S$

hence $Diam \ \bar{S} ≤ 2ε + Diam \ S$

since ε is arbitrary, the result is proven.

So, why do you think the proof fails for isolated points?? Where did we use that points were not isolated?

 

[Bachelier]

HallsofIvy wrote:

"...The diameter of set A, as well as the diameter of its closure, depends upon the entire set, not individual points.

If, for example, $A=(0,1)∪ \{2\}$then,since we have points arbitrarily close to 0 in the set,the diameter of A is 2 − 0= 2.
The closure of A is,of course,

$[0,1]∪\{2\}$ which still has diameter 2.
Another example is $A=(0,1)∪\{2\}∪(3,4)$ which has diameter 4 - 0= 4. It's closure is $[0,1] ∪ \{2\} ∪ [3,4]$ which still has diameter 4."

micromass wrote:
So, why do you think the proof fails for isolated points?? Where did we use that points were not isolated?

You know what confused me is the fact that I forgot that the isolated point would be part of $S$ in the first place.

We define the boundary of a set as being the limit points of the set + isolated points of the set.

But an isolated point of a set can only be an element of the boundary if it is an element of the original set.

 

[Bachelier]

please see attached