# Thread: Showing an open 2-ball has an open cover that does not have a finite subcover

1. ## Showing an open 2-ball has an open cover that does not have a finite subcover

We are in $(\mathbb{R}^2, d)$, where $d$ is the Euclidean metric. Let our region be the open ball centered at the point $x=(-2,0)$ with radius 1, denoted by $B(x;1)$.

My attempt: Let $F_n= B(x;1-\frac{1}{n})$ where $n\in \mathbb{N}$.

It's obvious to me (not sure if I must prove) that $\{ F_n \}$ is an open cover for $B(x;1)$ since $B(x;1) \subset \bigcup_{n\in\mathbb{N}} F_n$

Suppose there exists a finite subcover of $\{ F_n \}$, say $\{F_{n_{i}}\}$. I must find an element in $t \in B(x;1)$ such that $t \notin \{F_{n_{i}}\}$ to reach a contradiction, right?

Since $\{F_{n_{i}}\}$ is a finite subcover we have that $B(x;1) \subset \bigcup_{n=1}^{n_i} F_{n_i}$. So every point in $B(x;1)$ is also in $B(x; 1-\frac{1}{n_i})$. Consider $n_i + 1$. It's obvious that $1-\frac{1}{n_i} < 1 - \frac{1}{n_i+1} < 1$, so
$B(x; 1-\frac{1}{n_i + 1}) \not\subset B(x;1-\frac{1}{n_i})$since the radius is bigger, but
$B(x; 1-\frac{1}{n_i + 1}) \subset B(x;1)$ since the radius is still less than 1. We have reached a contradiction.

Is this a correct?

2. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

We are in $(\mathbb{R}^2, d)$, where $d$ is the Euclidean metric. Let our region be the open ball centered at the point $x=(-2,0)$ with radius 1, denoted by $B(x;1)$.
My attempt: Let $F_n= B(x;1-\frac{1}{n})$ where $n\in \mathbb{N}$.

It's obvious to me (not sure if I must prove) that $\{ F_n \}$ is an open cover for $B(x;1)$ since $B(x;1) \subset \bigcup_{n\in\mathbb{N}} F_n$

Suppose there exists a finite subcover of $\{ F_n \}$, say $\{F_{n_{i}}\}$. I must find an element in $t \in B(x;1)$ such that $t \notin \{F_{n_{i}}\}$ to reach a contradiction, right?

Since $\{F_{n_{i}}\}$ is a finite subcover we have that $B(x;1) \subset \bigcup_{n=1}^{n_i} F_{n_i}$. So every point in $B(x;1)$ is also in $B(x; 1-\frac{1}{n_i})$. Consider $n_i + 1$. It's obvious that $1-\frac{1}{n_i} < 1 - \frac{1}{n_i+1} < 1$, so
$B(x; 1-\frac{1}{n_i + 1}) \not\subset B(x;1-\frac{1}{n_i})$since the radius is bigger, but
$B(x; 1-\frac{1}{n_i + 1}) \subset B(x;1)$ since the radius is still less than 1. We have reached a contradiction.
I do not see the contradiction. But then I not sure I follow the argument.
The collection $\{\mathscr{F}_n\}$ is known as a telescope. For each $m>n$ we have $\mathscr{F}_n\subset\mathscr{F}_m$.
Note that is a proper subset.
So where did you have a finite subcollection? If there a finite collection, then there is a "largest" $\mathscr{F}_m$, WHY?

Say that is $\mathscr{F}_k$ which does not cover $\mathscr{B}(x;1)$. WHY is that true????

What exactly is it that you are trying to prove?

3. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

Originally Posted by Plato
So where did you have a finite subcollection? If there a finite collection, then there is a "largest" $\mathscr{F}_m$, WHY?
My claim is that I cannot find a finite subcollection. For the second question, if a collection of open subsets covers a space and you can take a finite number of those subsets in that collection to cover that space, then that would be the largest $\mathscr{F}_m$ needed to cover the space.

Originally Posted by Plato
Say that is $\mathscr{F}_k$ which does not cover $\mathscr{B}(x;1)$. WHY is that true????
Because $\mathscr{F}_k \subset \mathscr{F}_{k+1}$, and $\mathscr{F}_{k+1}$ does not cover $\mathscr{B}(x;1)$.

Originally Posted by Plato
What exactly is it that you are trying to prove?
I'm trying to prove that the open cover I created does not have a finite subcover.

4. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

My claim is that I cannot find a finite subcollection. For the second question, if a collection of open subsets covers a space and you can take a finite number of those subsets in that collection to cover that space, then that would be the largest $\mathscr{F}_m$ needed to cover the space.

Because $\mathscr{F}_k \subset \mathscr{F}_{k+1}$, and $\mathscr{F}_{k+1}$ does not cover $\mathscr{B}(x;1)$.
Since any finite collection is contained in $\mathscr{F}_{N}$ where N is the largest k, what you have said is that these do **not** cover $\mathscr{B}(x;1)$.

I'm trying to prove that the open cover I created does not have a finite subcover.
You are trying to prove $\mathscr{B}(x;1)$ is not *compact*?

5. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

My claim is that I cannot find a finite subcollection. For the second question, if a collection of open subsets covers a space and you can take a finite number of those subsets in that collection to cover that space, then that would be the largest $\mathscr{F}_m$ needed to cover the space.

Because $\mathscr{F}_k \subset \mathscr{F}_{k+1}$, and $\mathscr{F}_{k+1}$ does not cover $\mathscr{B}(x;1)$.

I'm trying to prove that the open cover I created does not have a finite subcover.
So show that the $\mathscr{F}_k$ with the largest subscript cannot cover the whole set.

6. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

Originally Posted by HallsofIvy
You are trying to prove $\mathscr{B}(x;1)$ is not *compact*?
Yes, but I am having trouble writing arguing clearly that the open cover of $\mathscr{B}(x;1)$ I constructed does not contain a finite subcover.

7. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

Originally Posted by Plato
So show that the $\mathscr{F}_k$ with the largest subscript cannot cover the whole set.
To do that I must show that $\mathscr{F}_k \subset \mathscr{F}_{k+1}$, but $\mathscr{B}(x;1) \not\subset \mathscr{F}_{k+1}$?

Then I'd just use the fact that $k+1 > k$ and manipulate that to get that $1 - \frac{1}{k} < 1 - \frac{1}{k+1} < 1$

8. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

To do that I must show that $\mathscr{F}_k \subset \mathscr{F}_{k+1}$, but $\mathscr{B}(x;1) \not\subset \mathscr{F}_{k+1}$?
Then I'd just use the fact that $k+1 > k$ and manipulate that to get that $1 - \frac{1}{k} < 1 - \frac{1}{k+1} < 1$
I think that this is a case that notation gets in the way.
If ${B_n} = \left( {\frac{1}{n},1 + \frac{1}{n}} \right)$ then $\bigcup\limits_{n \in {Z^ + }} {{B_n}} = \left( {0,2} \right)$.

Can you prove that no finite collection of the $B_n$ covers $(0,2)~?$

9. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

Originally Posted by Plato
I think that this is a case that notation gets in the way.
If ${B_n} = \left( {\frac{1}{n},1 + \frac{1}{n}} \right)$ then $\bigcup\limits_{n \in {Z^ + }} {{B_n}} = \left( {0,2} \right)$.

Can you prove that no finite collection of the $B_n$ covers $(0,2)~?$
Let $\mathscr{B} = \{B_{n_1}, \ldots , B_{n_k} \}$ be some finite collection of $\{B_n:n\in \mathbb{N}\}$. Let m be the largest subscript, then $$\bigcup_{i=1}^k B_{n_i} = B_m = \left(\frac{1}{m}, 1+\frac{1}{m}\right).$$ Then it follows that the finite collection does not cover (0,2)

10. ## Re: Showing an open 2-ball has an open cover that does not have a finite subcover

Let $\mathscr{B} = \{B_{n_1}, \ldots , B_{n_k} \}$ be some finite collection of $\{B_n:n\in \mathbb{N}\}$. Let m be the largest subscript, then $$\bigcup_{i=1}^k B_{n_i} = B_m = \left(\frac{1}{m}, 1+\frac{1}{m}\right).$$ Then it follows that the finite collection does not cover (0,2)
Can you prove that collection cannot cover $(0,2)~?$
Can you prove that collection cannot cover $(0,2)~?$
Consider $\frac{1}{m+1} \notin \left(\frac{1}{m} , 1+ \frac{1}{m}\right)$. But $\frac{1}{m+1} \in (0,2)$ Thus that collection cannot cover (0,2)