# Injective function $f_k:\mathcal{P}_k(\mathbb{N})\to \mathbb{N}^k$ for $k\geq 1$ - just a mapping?

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.

• In your mind, is it $f_2(\{x\in\Bbb N\,:\, x^2-3x+2=0\})=(1,2)$ or is it $f_2(\{x\in\Bbb N\,:\, x^2-3x+2=0\})=(2,1)$? – Saucy O'Path Sep 30 '18 at 9:08
• 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 $f_k(A)$ – Holo Sep 30 '18 at 9:19
• @SaucyO'Path It's the same? :) EDIT: I see the point :) – Frederik Sep 30 '18 at 9:19
• @Holo Thank you! So $A=\{x_1 < x_2 < ... < x_k\}$? – Frederik Sep 30 '18 at 9:20
• @Frederik you need to explain why you know that the elements of $A$ and be written like this, but this works – Holo Sep 30 '18 at 9:22

Let a$$_1$$,.. a$$_k$$ be a well ordering of A and map A to (a$$_1$$,.. a$$_k$$).