hamming metric

Metric_Space

yes, so one of x_k is 0 then?

micromass

So, did you figure out why

[tex]\sum_{k=0}^{+\infty}{\frac{x_k}{2^k}}<1[/tex]

if one of the x_k is 0?

Metric_Space

Isn't it because that means the sum above is the difference of two other sums?

micromass

I'm not sure what you mean. Which other sums?

Metric_Space

one that sums to one, and one that has a zero at the kth entry?

micromass

Yes, indeed!

So, now we have shown that only (1,1,1,...) has the property that

[tex]\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}=1[/tex].

So, can you now describe the ball around (0,0,0,...) with radius 1?

Metric_Space

Would it be all entries are 1? But it wouldn't be finite...would it?

micromass

No, this isn't correct. I want you to find the set

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

So...

Metric_Space

Isn't that just all the (x_1,x_2, ...) minus the entry with (1,1,1,1...)?

micromass

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 [itex](a_1,a_2,a_3,...)[/itex] with radius 1?

Metric_Space

would it be everything but (a_1,a_2, a_3, ...)?

micromass

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...

Metric_Space

...I think I'll need another hint

micromass

For what sequence [itex](x_n)_n[/itex], does

[tex]\sum_{k=1}^{+\infty}{\frac{|x_k-a_k|}{2^k}}=1[/tex]

It's basically the same as my last question (when the a_k were simply 0). What must hold for |x_k-a_k| in order for the above series to be 1?

Metric_Space

|X_k-a_k| --> 0 as k--> infinity?

micromass

No. I don't know how you got that?
You do know that [itex]\frac{1}{2^k}|X_k-a_k|\rightarrow 0[/itex], but I fail to see how it is relevant here...

Metric_Space

Oh...does |X_k-a_k|=1 for k=1..infinity?