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anonymous
 one year ago
please i need help here.
anonymous
 one year ago
please i need help here.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X ε Rn. Show that the set B(X, ε ) is open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jtvatsim , @Michele_Laino and @zzr0ck3r

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\large B\left( {X,\varepsilon } \right) = \left\{ {\left( {{x_1},{x_2},...,{x_n}} \right):x_1^2 + x_2^2 + ... + x_n^2 < {\varepsilon ^2}} \right\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that all or this sir?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! That's all!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I have supposed that we can use the Euclidean metrics

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we can say that every sequence of Cauchy in X is convergent to each element of X

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if we consider any convergent sequence in X, then that sequence is also a Cauchy sequence, so the limit point belongs to X itself

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in that case I have at least O_n which is not empty since X is not empty by hypothesis

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2more precisely for each element of \[\Large {O_n}\] its closure is equal to X

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2another 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sorry, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://nou.edu.ng/uploads/NOUN_OCL/pdf/edited_pdf3/MTH%20301.pdf

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I 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"

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so O_n can not the empty set, otherwise we can not take any convergent sequence which converge to any element of X

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. apply that theorem and prove is sir

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since by hypothesis X is not empty

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as I said before, the intersection of the elements of O_n is a dense subet of X, so we can write: \[\Large \begin{gathered} {O_n} = \cup {0_n} \hfill \\ \hfill \\ \overline { \cap {0_n}} = X \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and of course: \[\Large X \ne \emptyset \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sorry 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so 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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Your 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..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please don't @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@GIL.ojei the first question is proved and here is why: our nsphere 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 nsphere 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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Just 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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Just 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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wish i never asked any questions that day. am so sorry guys. please forgive

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@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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