Convergence in Cantor space I have a sequence, in the Cantor space wrt the Cantor metric,$\{ a_{n}\} $   with the first n digits are 0 followed by 1s
Does this sequence converge?
Is it not convergent as there are lots of choices for  a possible limit?
Eg A= (1, 0, 0, 0, 0...) or B=(0, 1, 1, 1,....)
Could be limits as the distance between them and $ a_{N} $   is $2^{-N} $  which tends to 0 as N increases?
Not very keyed up on this as   cant find example in book or web of convergence                        
 A: If d(a,b) = $2^{-n}$ with n the first place the sequences differ, then for N > 0, the first place A and $a_N$ differ is the zeroth term (assuming it's zero indexed). For B, if N > 1, the first place is the first term. So I don't see how you think the distance goes to zero; for A the distance is always 1, and for B it's always $\frac{1}{2}$. On the other hand, if we define L = (0,0,0,...), then L and $a_N$ differ on the N+1 th term, so the distance is ${2}^{-(N+1)}$, which does go to zero, so $a_n$ converges to L.
A: If we have a sequence $a_n = (0,0,0,\ldots, 1, 1, 1, \ldots )$ where $a_n$ has $n$ zeroes, then this sequence converges to $(0,0,0,0,0,\ldots)$, as it does so 
in every coordinate: in a fixed coordinate $n$ the sequence is all $1$'s and then from the $n+1$-th term we only have $0$'s, so all tails are $0$'s in this $n$-th coordinate. So the $n$-th coordinate of the limit must also be $0$ (as projections are continuous). In the product topology convergence of a sequence is determined by its coordinate sequences. Hence my conclusion: the limit is $0$.
It is also clear if we look at the standard homeomorphism with the standard Cantor set in $[0,1]$: then $a_n$ maps to $\sum_{i=n+1}^\infty (\frac{2}{3})^{i} = 3(\frac{2}{3})^{n+1} \to 0 \,(n \to \infty)$.
Also in the Cantor metric: $d(a_n, 0) = 2^{-(n+1)} \to 0$ yielding the same conclusion again.
