## hamming metric

yes, so one of x_k is 0 then?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus So, did you figure out why $$\sum_{k=0}^{+\infty}{\frac{x_k}{2^k}}<1$$ if one of the xk is 0?
 Isn't it because that means the sum above is the difference of two other sums?

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 Quote by Metric_Space Isn't it because that means the sum above is the difference of two other sums?
I'm not sure what you mean. Which other sums?
 one that sums to one, and one that has a zero at the kth entry?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, indeed! So, now we have shown that only (1,1,1,...) has the property that $$\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}=1$$. So, can you now describe the ball around (0,0,0,...) with radius 1?
 Would it be all entries are 1? But it wouldn't be finite...would it?

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 Quote by Metric_Space Would it be all entries are 1? But it wouldn't be finite...would it?
No, this isn't correct. I want you to find the set

$$\left\{(x_1,x_2,x_3,...)~\mid~\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}<1\r ight\}$$

So...
 Isn't that just all the (x_1,x_2, ...) minus the entry with (1,1,1,1...)?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Indeed, so the ball around (0,0,0,...) with radius 1 is everything except (1,1,1,1,...). So, now the obvious generalization: what is the ball around $(a_1,a_2,a_3,...)$ with radius 1?
 would it be everything but (a_1,a_2, a_3, ...)?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Uuh, no, I get the impression that you're just guessing here. Obviously a=(a1,a2,a3,...) will be in the ball, since d(a,a)=0<1. Moreover, we have just calculated that the ball around (0,0,0,...) is everything but (1,1,1,...). Thus (0,0,0,...) is in the ball...
 ...I think I'll need another hint
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus For what sequence $(x_n)_n$, does $$\sum_{k=1}^{+\infty}{\frac{|x_k-a_k|}{2^k}}=1$$ It's basically the same as my last question (when the ak were simply 0). What must hold for |xk-ak| in order for the above series to be 1?
 |X_k-a_k| --> 0 as k--> infinity?

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You do know that $\frac{1}{2^k}|X_k-a_k|\rightarrow 0$, but I fail to see how it is relevant here...