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Apologies for the simple question but my google-fu has left me and my mind is weak...

Can all arbitrary sequences of decimal digits be put into one to one correspondence with the real numbers? How?

Decimal expansion won't work because 1 = 0.999999'

What about sequences of Integers or natural numbers?

2 Answers2

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Concerning your first question, since there is only countably many real numbers which admits two differents decimal expensions, you can indeed map $\{0,\dots,9\}^{\mathbb N}$ in one to one correspondance with $\mathbb R$.

For the second question, if you have a sequence of integers, then write each one in base $9$ and consider the real number which decimals in base $10$ are given by the numbers of your sequence separated by a digit $9$. You get a map which is not surjective but it is injective and you have an injective map in the other direction so by Cantor-Bernstein there exists a bijection.

Régis
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  • Got it. The pairs of equal decimal expansions are all rational and so countable. A possible mapping would be to double all the rationals and squeeze in the extra members of the pairs that way. Thanks, this scratched the itch in my mind. – Christopher Edwards Apr 24 '17 at 23:24
  • Yes, or otherwise, each time you have a sequence which, from a certain rank, has form 9, then k digits zeros, then 9 again, the $k$ digits 0,eerosetc ad infinitum, you replace it with the same one but the 9 being separated by $k+1$ zeros, then you take decimal expansion. – Régis Apr 24 '17 at 23:30
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The real numbers can be mapped* onto $(0,1)$ but not $[0,1]$. If you admit $0$ and $1 = 0.9999\dots$ as decimal expansions you need to include $\pm \infty$ with the reals to form the extended real number system.

*Here we are presuming that you want every real to have at least one image and that the mapping should be as close to 1-1 as possible.

JMJ
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