- anonymous

please i need help here.

- katieb

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- anonymous

Let X ε Rn. Show that the set B(X, ε ) is open

- anonymous

- Michele_Laino

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

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## More answers

- Michele_Laino

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

- anonymous

is that all or this sir?

- Michele_Laino

yes! That's all!

- anonymous

Let 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

I have supposed that we can use the Euclidean metrics

- Michele_Laino

we can say that every sequence of Cauchy in X is convergent to each element of X

- Michele_Laino

if 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

in that case I have at least O_n which is not empty since X is not empty by hypothesis

- Michele_Laino

more precisely for each element of \[\Large {O_n}\]
its closure is equal to X

- Michele_Laino

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

- anonymous

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?

- Michele_Laino

ok!

- anonymous

http://www.nou.edu.ng/uploads/NOUN_OCL/pdf/edited_pdf3/MTH

- anonymous

page 26 thorem

- Michele_Laino

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

- anonymous

http://nou.edu.ng/uploads/NOUN_OCL/pdf/edited_pdf3/MTH%20301.pdf

- anonymous

try that link

- Michele_Laino

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"

- Michele_Laino

space*

- Michele_Laino

so O_n can not the empty set, otherwise we can not take any convergent sequence which converge to any element of X

- anonymous

ok. apply that theorem and prove is sir

- Michele_Laino

since by hypothesis X is not empty

- Michele_Laino

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

- Michele_Laino

and of course:
\[\Large X \ne \emptyset \]

- Michele_Laino

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

- anonymous

ok

- Michele_Laino

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

- Michele_Laino

that's all!

- anonymous

##### 1 Attachment

- zzr0ck3r

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..

- anonymous

please don't @Michele_Laino

- Michele_Laino

@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.

- zzr0ck3r

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

- zzr0ck3r

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

- zzr0ck3r

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.

- anonymous

i wish i never asked any questions that day. am so sorry guys. please forgive

- zzr0ck3r

lol

- Michele_Laino

@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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