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

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

    Quote Originally Posted by MadSoulz View Post
    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?
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by Plato View Post
    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.

    Quote Originally Posted by Plato View Post
    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 $\mathscr{B}(x;1)$.

    Quote Originally Posted by Plato View Post
    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.
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by MadSoulz View Post
    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*?
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by MadSoulz View Post
    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.
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by HallsofIvy View Post
    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.
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by Plato View Post
    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 $1 - \frac{1}{k} < 1 - \frac{1}{k+1} < 1$
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by MadSoulz View Post
    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 $(0,2)~?$
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by Plato View Post
    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)
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by MadSoulz View Post
    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 $(0,2)~?$
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    Re: Showing an open 2-ball has an open cover that does not have a finite subcover

    Quote Originally Posted by Plato View Post
    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)
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