# Is there a homeomorphism from the space of sequences to [0,1]?

If I consider the norm for the space of sequences of digits {0-9} to mimic the norm for real numbers.

$|\left\{x_n\right\}| = \sum_{n=1}^{\infty} \frac{x_n}{10^n}$

shouldn't I now have a space identical to [0,1]?

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What kind of sequences are you considering? Are you looking at sequences with values in $0,\dots,9$ and not infinitely many repeating nines? – Olivier Bégassat Jan 8 '13 at 13:00
I'm not sure. If I exclude trailing 9's, shouldn't I exclude trailing 9's from [0,1] also? – user54358 Jan 8 '13 at 13:02
It might be necessary to exclude them though to get one-to-one... so yeah but if I stick with insisting it's a norm and just say that trailing 9's is equal to another sequence? Wouldn't that work just as well? – user54358 Jan 8 '13 at 13:11
If $x_n$ are restricted to integers only then there is a bijection from the sequence space to R. See here: math.stackexchange.com/questions/226723/…. To get a bijection from R to [0,1] see this trick here: answers.yahoo.com/question/index?qid=20080815000752AAZTABY. Finding a homeomorphism is more difficult of course, and although I am no expert, it looks to me that there is no continuous bijection between [0,1] and R, and so you cannot use my results very usefully... – Adam Rubinson Jan 8 '13 at 13:13
wouldn't the norm induce the same metric as in real numbers? – user54358 Jan 8 '13 at 13:25

Yes in this case.

By Cantor diagonal methold ( or Dedekind cut ) we can identify $\mathbb{R}$ and space of sequence $\{x_n\}_{n\in\mathbb{N}}$ in $\{0,1,2,3,4,5,6,7,8,9\}^\mathbb{N}$. By $[0,1]\ni x \mapsto \tan(x\cdot \pi)\in\mathbb{R}$ we cam identify $\{0,1,2,3,4,5,6,7,8,9\}^\mathbb{N}$ and $[0,1]$

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So the norm/metric should be defined by: $|\left\{x_n\right\}| = \sum_{n=1}^{\infty} \frac{x_n}{10^n}$ (provided that $\sum_{n=1}^{\infty} \frac{x_n}{10^n}$ exists) – Adam Rubinson Jan 8 '13 at 13:26
@AdamRubinson I update my answer. – MathOverview Jan 8 '13 at 13:51
Your function is not continuous. I think you meant f(x) = tan((x-1/2)pi), in which case I think the only problem is sequences like (0,9,9,9,9,9,...) and (1,0,0,0,0,...) which are not the equal, although 0.999... and 1 are equal and so injectivity is the problem now. – Adam Rubinson Jan 8 '13 at 14:02