# Diameter of the closure

1. Mar 5, 2013

### 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.

Last edited: Mar 5, 2013
2. Mar 5, 2013

### micromass

Staff Emeritus
Could you post the proof that Rudin gives??

3. Mar 5, 2013

### Bachelier

$Diam \ S ≤ Diam \ \bar{S}$ 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.

4. Mar 5, 2013

### HallsofIvy

Staff Emeritus
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.

5. Mar 5, 2013

### micromass

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

6. Mar 5, 2013

### Bachelier

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.

Last edited: Mar 5, 2013
7. Mar 5, 2013