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Closure of open ball

  1. Apr 23, 2016 #1
    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. jcsd
  3. Apr 23, 2016 #2

    Samy_A

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    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
  4. Apr 23, 2016 #3
    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.
     
  5. Apr 23, 2016 #4

    Samy_A

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    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
  6. Apr 24, 2016 #5
    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.
     
  7. Apr 24, 2016 #6
    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.
     
  8. Apr 24, 2016 #7

    Samy_A

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    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##.
     
  9. Apr 24, 2016 #8
    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!
     
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