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I have to find an injective function $f_k:\mathcal{P}_k(\mathbb{N})\to \mathbb{N}^k$ for $k\geq1$, where $\mathcal{P}_k(\mathbb{N})$ is the power set of $\mathbb{N}$ with k elements.

I have trouble understanding such a function, as I only can come with: Given a set $A=\{x_1,x_2,...,x_k\}\in\mathcal{P}_k(\mathbb{N})$ with $k$ natural numbers, then $f_k(A)=(x_1,x_2,...,x_k)\in\mathbb{N}^k$. So it is a function that takes $k$ natural numbers and gives a vector/point in the room of natural numbers with orden $k$?

Furthermore, such a function must be injective: When $A_1, A_2, A_3,...$ be sets with $k$ natural numbers, then there union must be countable. So the cardinality of $\#\mathcal{P}_k(\mathbb{N})\leq\#\mathbb{N}$, as this is the definition of countability. When $\#\mathcal{P}_k(\mathbb{N})\leq\#\mathbb{N}$ the function $f$ must be injective.

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    $\begingroup$ In your mind, is it $f_2(\{x\in\Bbb N\,:\, x^2-3x+2=0\})=(1,2)$ or is it $f_2(\{x\in\Bbb N\,:\, x^2-3x+2=0\})=(2,1)$? $\endgroup$ – Saucy O'Path Sep 30 '18 at 9:08
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    $\begingroup$ The problem with your function is that in $A$ there is no order and in $f_k(A)$ you have order, you need to be specific about how you order them in $f_k(A)$ $\endgroup$ – Holo Sep 30 '18 at 9:19
  • $\begingroup$ @SaucyO'Path It's the same? :) EDIT: I see the point :) $\endgroup$ – Frederik Sep 30 '18 at 9:19
  • $\begingroup$ @Holo Thank you! So $A=\{x_1 < x_2 < ... < x_k\}$? $\endgroup$ – Frederik Sep 30 '18 at 9:20
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    $\begingroup$ @Frederik you need to explain why you know that the elements of $A$ and be written like this, but this works $\endgroup$ – Holo Sep 30 '18 at 9:22
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Let a$_1$,.. a$_k$ be a well ordering of A and map A to (a$_1$,.. a$_k$).

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