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The power set $\mathcal{P}(X) = \text{Map}(X, \{0, 1\})$. Hence, $|\text{Map}(\Bbb N, Y)| = |\Bbb R|$ where $|Y|=2$. Prove for any finite set $Y$, the equality still holds.

I know I can map the powerset to the real numbers by considering an element in the powerset of naturals and using the subset as a way to create a decimal expansion, reversing it with a binary expression to get the backwards case.

Formally, an element $A\in \mathcal{P}(\mathbb{N})$ is a subset of the naturals. So, $$f(A)=\sum_{k\in A}\frac{1}{10^k}:\mathcal{P}(\mathbb{N})\rightarrow[0, 1]$$

The reverse is:

$$g(x)=\{i:n_i=1\}\subset \mathbb{N}:(0,1)\rightarrow\mathcal{P}(\mathbb{N}).$$

I wonder if I can do something similar for the question here where we kind of extend the powerset from a 2 to a 3 and onwards

Turquoise Tilt
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Edward Cheung
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1 Answers1

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Your question is rather hard to follow and has several inconsistencies or outright mistakes. First of all, the base you want to use for $f$ is $2,$ not $10$ since the only digits that are at your disposal are $0$ and $1.$ Moreover, note that the function described as $$ f : \mathcal{P}(\mathbb{N}) \to [0, 1], f(A) := \sum_{k \in A} \frac{1}{2^k} $$ is not actually a bijection since, for example, the subsets $A_1 := \{ 1 \}$ and $A_2 := \{2, 3, 4, \dots\}$ are mapped to the same element of $[0, 1],$ namely $\frac{1}{2}.$ However, you are able to easily fix this problem, see for example this post on MSE.

In general I believe you are asking about functions defined as $f : \mathbb{N} \to \{0, 1, \dots, n - 1\} =: \mathfrak{n}.$ It is indeed possible to extend the result and it is not hard to see why. I propose it to you as an exercise to find a function $e:\mathfrak{n}^{\mathbb{N}} \to [0, 1]$ defined in a similar way to the one in the post from before that is a bijection between the two sets and is basically the expansion of a number in $[0, 1]$ in base $n,$ i.e. for most of the sequences in $\mathfrak{n}^{\mathbb{N}}$ you will have that $$e((a_j)_{j \in \mathbb{N}}) = \sum_{j \in \mathbb{N}} \frac{a_j}{n^j}.$$ For the rest of the numbers, you need to do something similar to what the accepted answer author did with the $p_n'$s and $q_n'$s. I hope this helps. :)

Actually Fritz
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