I don't understand how the diameters of $ \bar A$ and $A$ are equal

Question

I'm given the assignment to prove that $\operatorname{diam}\bar A = \operatorname{diam}A$, where $\operatorname{diam}A=\sup\{\rho(a,b): a,b \in A\}.$

How can they be equal if $\bar A= \partial A \cup A$:

Say $x\in \partial A$, then does it not follow that $x > a, \; x> b$, thus $ x \notin A$ ?

What is$\rho(a,b)$ ? Are you sure that the set A is defined as it is ? — Mohammad Riazi-Kermani, Mar 18 '18 at 14:06
I have guessed that you're working on a metric space and $\rho$ is the given metric. I also fixed your definition, adding “diam” and “sup”. — egreg, Mar 18 '18 at 14:09
Because it is the supremum. $\mbox{diam } A=\sup {d(a,b):a,b\in A}$. — Dog_69, Mar 18 '18 at 14:10
@Mohammad it is distance between two points in metric space. I suppose it is defined as d(a,b) commonly. — flowian, Mar 18 '18 at 14:10
@dog isn't supremum in this case greatest distance between points a and b in A? — flowian, Mar 18 '18 at 14:14
If $a,b,x$ are points in a metric space, then what could it mean to say $x>a$ or $x>b \text{ ?} \qquad$ — Michael Hardy, Mar 18 '18 at 15:51

score 1 · Accepted Answer · answered Mar 18 '18 at 14:12

1

Consider the intervals $(0,2)$ and $[0,2]$. One is open -- not containing either $0$ or $2$. The other is closed -- containing both $0$ and $2$.

The length of both these intervals is $2-0 = 2.$

answered Mar 18 '18 at 14:12

Eff

12,989

That actually makes sense. Thank you – flowian Mar 18 '18 at 14:17

score 1 · Answer 2 · answered Mar 18 '18 at 14:19

Since $A\subseteq\bar{A}$, it is obvious that $\operatorname{diam}A\le\operatorname{diam}\bar{A}$.

If $\operatorname{diam}A=\infty$, there is nothing to prove, so we can assume $\operatorname{diam}A$ is finite.

Suppose $\operatorname{diam}\bar{A}>\operatorname{diam}A$. Then there are $x,y\in\bar A$ such that $\rho(x,y)>\operatorname{diam}A$.

Let $\varepsilon=(\rho(x,y)-\operatorname{diam}A)/2$. Then, since $x,y\in\bar{A}$, there are $a,b\in A$ such that $\rho(a,x)<\varepsilon$ and $\rho(b,y)<\varepsilon$. By applying the triangle inequality we have $$ \rho(x,y) \le\rho(x,a)+\rho(a,y) \le\rho(x,a)+\rho(a,b)+\rho(b,y)< \varepsilon+\operatorname{diam}A+\varepsilon<\rho(x,y) $$

score 0 · Answer 3 · answered Mar 18 '18 at 14:19

We don't necessarily have any ordering among the points of the metric space (consider e.g. $\Bbb R^n$), so your statements $x>a$ and $x>b$ are meaningless.

To prove the claim, take any $a, b\in\bar A$ and consider sequences $a_n, b_n\in A$ with $a_n\to a, \ b_n\to b$, and note that $\rho(a_n, b_n) \to\rho(a, b) $.

score 0 · Answer 4 · answered Mar 18 '18 at 14:32

0

You want to show $$\operatorname{diam}\bar A = \operatorname{diam}A$$

Obviously $$\operatorname{diam}A \le\operatorname{diam}\bar A $$

On the other hand if $$\operatorname{diam}A <\operatorname{diam}\bar A $$ we have two points in the $\bar A$ whose distance is greater than $\operatorname{diam}A$

At least one of these points must be in $\partial A $

That implies that the point in the boundary is included in an open ball which does not intersect A.

That contradicts the definition of $\partial A $

answered Mar 18 '18 at 14:32

Mohammad Riazi-Kermani

68,728

That implies that the point in the boundary is included in an open ball which does not intersect A.

How does it imply that open ball of the point in $\partial A$ does not intersect $A$ ?
– flowian Mar 19 '18 at 11:13
The distance between two points is greater than the diameter so one of these points, say $b$ is on the boundary. The distance minus the diameter is $ r>0$. The open ball B(b;r/2) does not intersect A, due to the triangle inequlity. – Mohammad Riazi-Kermani Mar 19 '18 at 12:38

I don't understand how the diameters of $ \bar A$ and $A$ are equal

4 Answers4