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Let $^\mathbb{N}\{0,1\}$ be defined as $\{f~|~ f:\mathbb{N}\to\{0,1\}\}$, i.e., all the functions that map $\mathbb{N}$ to the set $\{0,1\}$. We want to show that this set is in bijection with $\mathbb{R}$.

I've been having trouble figuring this out. If we can show there is an injection from $^\mathbb{N}\{0,1\}\to\mathbb{R}$ and an injection $\mathbb{R}\to ~^\mathbb{N}\{0,1\}$, then we can invoke the Schroder-Bernstein theorem to complete the argument that the two sets are in bijection with one another.

We can represent each $f\in~^\mathbb{N}\{0,1\}$ as the dyadic expansion of a real number $r_j\in[0,2]$, i.e., $$ r_j = \sum_{k=0}^\infty \frac{f(k)}{2^k}$$ But $r_j$ may be mapped to by more than one function, so I am unsure as to how to proceed. Any help is appreciated.

Andrés E. Caicedo
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locallyeuclidean
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    An example: $\displaystyle\frac{1}{2}=\sum_{n=2}^\infty \frac{1}{2^n}$ – André Porto Oct 31 '18 at 03:19
  • But the number of points for which this ambiguity exists is countable, so you can list them in some order, and then by choosing for a given repetition, say, the least such in the order, you will obtain a bijection. – Matematleta Oct 31 '18 at 03:50

2 Answers2

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Yes there is a problem for functions $f$ that become constant equal to $1$ at $\infty$. So you should rather take $f \mapsto \sum_k \frac{f(k)}{3^k}$.

  • There is no function whose domain is $\Bbb N$ that is equal to anything at $\infty$. Specifically, since $\infty$ is not a natural number. – Asaf Karagila Oct 31 '18 at 07:56
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The function you have is injective on the set of non-terminating sequences. (i.e. sequence that have infinitely many $1$'s). So define $f$ the way you have done on non-terminating sequences. There are only countable number of terminating sequences and you can map them to $\{2+\frac 1 n:n=1,2,\cdots\}$ for example. This gives an injective map into $\mathbb R$.

Kavi Rama Murthy
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