# Closure of open ball

1. Apr 23, 2016

### Incand

1. The problem statement, all variables and given/known data
Show that $\overline{B(a,r)} = \{ x \in \mathbf R^n ; |a-x| \le r \}$ in $\mathbf R^n$ for all points $a \in \mathbf R^n$ and $r>0$.
Is it possible to generalize the statement to any normed vector space?
Give a example of a metric space where the statement is not true.

2. Relevant equations
Definition:
If $X$ is a matric space, if $E \subset X$, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $\bar E = E \cap E'$.

Definition:
Let $(M,d)$ be a metric space. An open ball with center in $\mathbf a \in M$ and radius $r>0$ is the set
$B(a,r) = \{x \in M;d(x,a)< r\}$.

3. The attempt at a solution
Starting with the first part:
In $\mathbf R^n$ the open ball becomes
$B(a,r) = \{x \in M; ||\mathbf a -\mathbf x|| < r\}$.
Since $\bar B$ is union of all limits point and the $B$ I need to show that an arbitrary point
$p\in \{x \in M; ||\mathbf a -\mathbf x|| = r\}$ is a limit point
while
$p\in \{x \in M; ||\mathbf a -\mathbf x|| > r\}$ isn't one.

Take any $p\in \{x \in M; ||\mathbf a -\mathbf x|| = r\}$. $p$ is a limit point of $B$ if every open ball to $p$ contains a point $q \ne p$ such that $q \in B$.
That is if for all $\epsilon > 0$ there is a point $q \in \{x\in M; ||p-x||<\epsilon \} \cap B$ (since $p$ and $q$ are from two disjoint sets).

This seems to be to be true since "for any $\epsilon > 0$ there is an infinite number of real numbers in the open ball" but I'm not sure what is expected of a proof or how to prove it.

2. Apr 23, 2016

### Samy_A

There is a typo in the "relevant equations": "the closure of $E$ is the set $\bar E = E \cap E'$" must be "the closure of $E$ is the set $\bar E = E \cup E'$.

To prove that $p\in \{x \in \mathbb R^n; ||\mathbf a -\mathbf x|| = r\}$ is a limit point , consider the following sequence $(1-\frac{1}{n})p$.

EDIT: that would work if $a=0$. For $a \neq 0$, just translate everything and consider the sequence $a + (p-a)(1-\frac{1}{n})$.

Last edited: Apr 23, 2016
3. Apr 23, 2016

### Incand

Been thinking over this a while and I'm not entirely sure what you meant with the sequence and how I should use it (limiting $\epsilon$ or as a point in $B$). I tried to prove the statement for the open ball centered around $\mathbf 0$ as a start

Any point q satisfying $q = (1-\frac{1}{n})r, \; \; n\in \mathbb Z^+$ is therefore in $B(\mathbf 0 ,r)$ since $(1-\frac{1}{n})r< r, \; \forall n \in \mathbb N$.

The sequence $(1-\frac{1}{n})r \to r$ as $n \to \infty$. This means that
$\forall \epsilon>0 \; \exists N \in \mathbb Z^+$ such that $|(1-\frac{1}{n})r-r|< \epsilon, \; \forall n \ge N$.
So for any $\epsilon$ the point $q$ satisfying $||q||=|(1-\frac{1}{n})r|$ for a sufficiently large $n$ is within the open ball $\{ x \in \mathbb R^n; ||\mathbf x - \mathbf p||< \epsilon\}$. Hence $\mathbf p$ is a limit point.

4. Apr 23, 2016

### Samy_A

Well, this is more or less what you had to do.
The formal argument I suggested with the sequence was (taking the case $a=0$, the case $a \neq$ is very similar):
For $p\in \{x \in \mathbb R^n; ||\mathbf 0 -\mathbf x|| = r\}$, the elements of the sequence $p(1-\frac{1}{n})$ lie in $B(0,r)$, and the sequence converges to $p$. You actually proved these two statements. It follows that $p$ lies in the closure of $B(0,r)$.

Last edited: Apr 23, 2016
5. Apr 24, 2016

### Incand

So you are using $p \in \mathbf R^n$ instead of the metric. I guess that would help with the last step I did, which felt shaky. Problem is I never worked with a sequence multiplied with an element from $\mathbf R^n$.
I didn't see your edit before now but I can see how using $p-\frac{p-a}{n}$ give the same result for an open ball around $a$.

For the next part:
To show that $p \in \{x \in \mathbb R^n; ||x||>r\}$ isn't a limit point to $B(a,r)$ take the open ball around $p$ with radius $\epsilon = \frac{||p-a||-r}{2}$. Then for a point $q$ in the open ball $||p-q||<\frac{||p-a||-r}{2} \le \frac{||p-q||+||q-a||-r}{2}$. Hence
$r\le ||p-q||+r < ||q-a||$. So $||q-a|| > r$ and $q$ isn't a limit point.

6. Apr 24, 2016

### Incand

As for the two last questions I believe I found a counterexample to the second one.
Take the discrete metric
$d(x,y) = \begin{cases}1 \; \text{ If } x \ne y\\ 0 \; \text{ If } x = y.\end{cases}$ and the ball $B(0,0.5)$. Then there is not points $q$ satisfying $d(q,a) = 0.5$ regardless of $a$ so $q$ can't be a limit point so this would be a counterexample.

I suspect the other question is true, that the statement can be generalised to arbitrary normed vector spaces. At least for the proof that the point $||q-a||>r$ if $||p-a||>r$ only uses the triangle inequality which is always true for normed vector spaces. The first part I did in euclidean space but using your approach I don't see anything making it invalid.

7. Apr 24, 2016

### Samy_A

Yes, this is correct, with two typos:
- $p \in \{x \in \mathbb R^n; ||x||>r\}$ must be $p \in \{x \in \mathbb R^n; ||x-a||>r\}$
- the conclusion is that $p$ isn't a limit point, since you proved that the open ball $p$ with radius $\epsilon$ is disjoint from $p \in \{x \in \mathbb R^n; ||x-a||<r\}$
Yes, the discrete metric is a good counterexample.
As far as general normed spaces are concerned, the proof is indeed valid for any norm. Nowhere did you use a specific property of $\mathbb R^n$.

8. Apr 24, 2016

### Incand

Thanks for all the help! I did the proof with the ball around 0 first and forgot to add the a when I realised I could do the same for any ball!