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Is this a valid argument about box topology? 
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#1
May212, 12:51 PM

P: 18

Given the following sequence in the product space R^ω, such that the coordinates of x_n are 1/n,
x1 = {1, 1, 1, ...} x2 = {1/2, 1/2, 1/2, ...} x3 = {1/3, 1/3, 1/3, ...} ... the basis in the box topology can be written as ∏(1/n, 1/n). However, as n becomes infinitely large, the basis converges to ∏(0, 0), which is a single set and not open. Therefore the sequence is not convergent in box topology. Since there exists a basis that converges to ∏(x, x), for any element x of R^ω, a sequence in box topology does not converge to any element in R^ω. 


#2
May212, 04:20 PM

Sci Advisor
P: 3,256

What kind of convergence are you talking about? Are you talking about a sequence of sets or a sequence of points? Under the usual definition of convergence, a sequence of points in a toplogical space that converges will converge to a point. There is no requirement that it converge to an open set. 


#3
May212, 05:44 PM

P: 18

I don't know the right terminology. x_n represents a point in R^ω that has infinite number of coordinates. I want to use the sequence that if each of the coordinate converges as n grows large, x_n converges to a point.
I want to show the difference between product topology and box topology using the sequence x_n = {1/n, 1/, 1/n, ..., 1/n, ... }. A textbook argument, if I read correctly, says that because 1/n eventually goes to 0, there is no open set in the box topology that contains (δ, δ) in R, hence no function converges. Am I on the right track? 


#4
May212, 09:28 PM

Sci Advisor
P: 3,256

Is this a valid argument about box topology?
To get a clear answer, you're going to have state a clear question. You mention a sequence of points and then you talk about a function converging without explaining what function you mean.
If there is a textbook argument, then quote the argument. Quote it, don't just give a mangled summary. (Perhaps the discipline of copying it will make it clearer to you.) 


#5
May212, 09:50 PM

P: 18

Thanks but no thank you. You are not being helpful at all.



#6
May212, 10:50 PM

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