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anonymous

  • one year ago

please i need help here.

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  1. anonymous
    • one year ago
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    Let X ε Rn. Show that the set B(X, ε ) is open

  2. anonymous
    • one year ago
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    @jtvatsim , @Michele_Laino and @zzr0ck3r

  3. Michele_Laino
    • one year ago
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    if X is a point of R^n and B(X,\epsilon) is an open ball centered at X whose radius is \epsilon >0, then that subset is the subsequent set: \[\Large B\left( {X,\varepsilon } \right) = \left\{ {\left( {{x_1},{x_2},...,{x_n}} \right):x_1^2 + x_2^2 + ... + x_n^2 < \varepsilon ^2 } \right\}\] namely it is an hypersphere whose radius is \epsilon, without its boundary, so it is open

  4. Michele_Laino
    • one year ago
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    \[\large B\left( {X,\varepsilon } \right) = \left\{ {\left( {{x_1},{x_2},...,{x_n}} \right):x_1^2 + x_2^2 + ... + x_n^2 < {\varepsilon ^2}} \right\}\]

  5. anonymous
    • one year ago
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    is that all or this sir?

  6. Michele_Laino
    • one year ago
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    yes! That's all!

  7. anonymous
    • one year ago
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    Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty

  8. Michele_Laino
    • one year ago
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    I have supposed that we can use the Euclidean metrics

  9. Michele_Laino
    • one year ago
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    we can say that every sequence of Cauchy in X is convergent to each element of X

  10. Michele_Laino
    • one year ago
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    if we consider any convergent sequence in X, then that sequence is also a Cauchy sequence, so the limit point belongs to X itself

  11. Michele_Laino
    • one year ago
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    in that case I have at least O_n which is not empty since X is not empty by hypothesis

  12. Michele_Laino
    • one year ago
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    more precisely for each element of \[\Large {O_n}\] its closure is equal to X

  13. Michele_Laino
    • one year ago
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    another valid statement can be this: any metric space , with the collection of all subsets that are open with respect to the metric of the space, is a topological space

  14. anonymous
    • one year ago
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    can i give you a link sir so that you check out the prove and see if it is correct and also make your own deductions prove from it?

  15. Michele_Laino
    • one year ago
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    ok!

  16. anonymous
    • one year ago
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    http://www.nou.edu.ng/uploads/NOUN_OCL/pdf/edited_pdf3/MTH

  17. anonymous
    • one year ago
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    page 26 thorem

  18. Michele_Laino
    • one year ago
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    sorry, I forgot, we can use the theorem of Category of Baire, so we can conclude that the intersection of the elements of O_n is dense in X, so the closure O_n is equal to X

  19. anonymous
    • one year ago
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    http://nou.edu.ng/uploads/NOUN_OCL/pdf/edited_pdf3/MTH%20301.pdf

  20. anonymous
    • one year ago
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    try that link

  21. Michele_Laino
    • one year ago
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    I don't know, I prefer to apply the Theorem of Category of Baire, in the subsequent form: "In a complete metric sopace, any countable collection of dense open subset has dense intersection"

  22. Michele_Laino
    • one year ago
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    space*

  23. Michele_Laino
    • one year ago
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    so O_n can not the empty set, otherwise we can not take any convergent sequence which converge to any element of X

  24. anonymous
    • one year ago
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    ok. apply that theorem and prove is sir

  25. Michele_Laino
    • one year ago
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    since by hypothesis X is not empty

  26. Michele_Laino
    • one year ago
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    as I said before, the intersection of the elements of O_n is a dense subet of X, so we can write: \[\Large \begin{gathered} {O_n} = \cup {0_n} \hfill \\ \hfill \\ \overline { \cap {0_n}} = X \hfill \\ \end{gathered} \]

  27. Michele_Laino
    • one year ago
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    and of course: \[\Large X \ne \emptyset \]

  28. Michele_Laino
    • one year ago
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    sorry befor I said an incorrect statement: "so the closure O_n is equal to X" I meant that the closure of the intersection of elements of O_n is equal to X

  29. anonymous
    • one year ago
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    ok

  30. Michele_Laino
    • one year ago
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    so if O_n is the empty set, then also the set: \[\Large { \cap {0_n}}\] is empty, and then since the closure of the empty set is the empty set it self, then X= empty set, which is false by hypothesis

  31. Michele_Laino
    • one year ago
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    that's all!

  32. anonymous
    • one year ago
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  33. zzr0ck3r
    • one year ago
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    Your first question was not proved by @michele_Laino. You cant just say it is open and thus it is. You must prove it. There is nothing nowhere that says that something without a border is open. This is NOT the definition of open set. There are a few methods you may use here depending on your definition of open 1) Show that it contains of all of its interior points and none of its closure 2) show that it is the arbitrary union of open sets 3).... There are many defiitions that work for open, but it was not proved here. You are getting math proofs from someone who studies physics and thinks pictures justify proof. they do not. Rigorous math is much different..

  34. anonymous
    • one year ago
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    please don't @Michele_Laino

  35. Michele_Laino
    • one year ago
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    @GIL.ojei the first question is proved and here is why: our n-sphere is a subset of R^n which is a manifolds, and, in particular, we can state that for any subset A of R^n, A is closed if and only if it contains all its boundary points, namely if and only if the subsequent expression holds: \[\Large B\left( {X,\varepsilon } \right) = Int\left\{ {B\left( {X,\varepsilon } \right)} \right\} \cup \partial \left\{ {B\left( {X,\varepsilon } \right)} \right\}\] where \[\Large \partial \] is the boundary of \[\Large B\left( {X,\varepsilon } \right)\] The proof of that expression is a simple exercise. Now, we can see that the closed n-sphere without its boundary is equal to its interior, which, by definition of interior of a set, it is open with respect to the subspace topology. Finally, just for information, when I was at university, I studied both mathematics and physics, in fact I wrote my degree thesis on partial differential equations and their symmetry operators, now such topics depend heavily on Manifolds, Lie algebras and vector spaces, in particular, so I know how to construct a rigorous proof.

  36. zzr0ck3r
    • one year ago
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    Just to let you know, when people are looking for proof in introduction, you should not use things that they done know about. There is no way that someone does not understand an open set understands what a manafold is. You have to know that what you showed does not help him at all. So silly...just silly

  37. zzr0ck3r
    • one year ago
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    Just to let you know, when people are looking for proof in introduction subjects, you should not use things that they don't know about. There is no way that someone does not understand an open set understands what a manafold is. You have to know that what you showed does not help him at all. So silly...just silly

  38. zzr0ck3r
    • one year ago
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    The fact that you have two mathematicians telling you that you are wrong on previous questions, and showed you but you still argue with no proof that you are right and just say topology over and over when it has nothing to do with the question, lets me know that your research..lets just say I would not read it or trust it.

  39. anonymous
    • one year ago
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    i wish i never asked any questions that day. am so sorry guys. please forgive

  40. zzr0ck3r
    • one year ago
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    lol

  41. Michele_Laino
    • one year ago
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    @GIL.ojei I'm sorry, for the previous proof, it is suffice to consider R^n a topological space, even if a manifold is also a topological space.

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