Proving U(d1,...,dp) is open and a basis for the Hamming metric

[Metric_Space]

around an arbitrary element...wouldn't it be similar to before.

ie. (sum,k=1..infinity, (x_k-a_k)/2^k) < (1/2)^n

either x_k is with 1's starting in the kth position and 0's afterwards
and a_k is ..not sureOR 1's in the (K+1)st position

and a_k is not sure

 

[micromass]

Well, the balls around a with radius 1/2 are either:

1) elements which agree with a in it's first coordinate
2) elements which agree with a except possibly in it's first coordinate.

Can you see why?
Can you extend this to balls with radius (1/2)^k??

 

[Metric_Space]

no...I don't follow

 

[micromass]

Can you see why it is true for a=0?

 

[Metric_Space]

I think so?

 

[micromass]

OK, npw apply the same reasoning to arbitrary a instead of 0...

 

[Metric_Space]

...I think I need a hint

 

[micromass]

Well, when does

$$\sum_{k=1}^{+\infty}{\frac{|x_k-a_k|}{2^k}}=\frac{1}{2}$$

??

 

[Metric_Space]

x_k=(1,0,0,0,0...) and a_k=(0,0,0,...)

or

x_k=(0,1,0,0...) and a_k=(0,0,0,...)

or

a_k=(1,0,0,0,0...) and x_k=(0,0,0,...)

a_k=(0,1,0,0...) and x_k=(0,0,0,...)

 

[micromass]

No, can you show my how you got that?

If

$$\sum_{k=1}^{+\infty}{\frac{|x_k-a_k|}{2^k}}=\frac{1}{2}$$,

then what must hold for $|x_k-a_k|$.

 

[Metric_Space]

I have a new question.

How would I show that the metric space defined by the Hamming metric is complete?