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The answer to this is given, but vaguely and the finishing details are left to be done. I unfortunately am at a loss of an idea as to how to finish it, so was wondering how one would do this last step, given in the instructions.

I am told to for every $$\{n_1,n_2,...,n_k\}\subseteq \mathbb N (n_1<n_2....<n_k) $$

assign

$$0.00..10001...10...1$$

where $1$ is are on places $n_1,....,n_k$ or fraction $$\frac{1}{2^{n_1}}+...\frac{1}{2^{n_k}}$$ and then to map this set to $\mathbb N$. My question is how do I map this set to $\mathbb N$ ??

Bozo Vulicevic
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  • Hint: use the Fundamental Theorem of Arithmetic. – Ángel Valencia Nov 30 '15 at 20:52
  • I think it's clearer if, instead of interpreting your binary string as a base $2$ "deci"mal , interpret it as a base $2$ integer. That makes the bijection more explicit. – lulu Nov 30 '15 at 20:56
  • The FTA would help to construct a bijection, but not to explain the bijection that we are given. – Justpassingby Nov 30 '15 at 20:56
  • Can someone put this together properly? – Bozo Vulicevic Nov 30 '15 at 20:57
  • For the record: I don't see this question as a duplicate of the other. The OP is asking about a particular bijection, not simply a proof of countability. – lulu Nov 30 '15 at 21:00
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    I agree with $lulu$. The OP is how to map those binary fractions to the natural numbers, not proving the countability of the finite subsets of the natural numbers. – Rory Daulton Nov 30 '15 at 23:15

1 Answers1

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Other than zero, which corresponds to the empty set, the resulting fractions from that procedure are those of the form

$$\frac{j}{2^k}$$

for some natural numbers $j,k$ and $j$ odd. Therefore you can list them out in this order:

$$\frac 1{2^1}$$ $$\frac 1{2^2},\frac 3{2^2}$$ $$\frac 1{2^3},\frac 3{2^3},\frac 5{2^3},\frac 7{2^3}$$

and so on: I'm sure you get the idea. All the binary fractions are included and there are no repetitions. It is not difficult to figure out a formula for that list: ask if you need one.

Since you ask, here is the formula:

$$\frac j{2^k} \mapsto \frac{2^k+j+1}2$$

Then zero (for the empty set) maps to $1$, $\dfrac 1{2^1}$ maps to $2$, and so on. (Note that I am using the version of $\Bbb N$ that does not include zero.)

That said, there is a better bijection from finite subsets of $\Bbb N$ to $\Bbb N$. Map any $A\subseteq\Bbb N$ to

$$1+\sum_{j\in A}2^{j-1}$$

In other words, if $A=\{n_1,n_2,\ldots,n_k\}$, then $A$ maps to one plus the number in binary notation

0.00..10001...10..1

$$1..01...10001..00$$

In other words, the same digits as your representation but in reverse order and before the implied decimal (binary?) point. Then $\{\}$ maps to $1$, $\{1\}$ maps to $2$, and so on.

Note that this bijection is for the "natural numbers" without zero. If zero is included, the bijection becomes easier:

$$A \mapsto \sum_{j\in A}2^j$$

Rory Daulton
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