I am having trouble showing that this is a partition:
$\{A_k|k\in \mathbb N \cup \{0\}\}$ where each $A_k=\{2^kn | n \in \mathbb N \text{ and n - odd}\}$ is a partition of the natural numbers.
I understand I need to use the definition of partition but I am not to sure how to apply this. For the first part of the definition I think I need to say something like X in Ak. I want to solve it using the three part of the definition of partitions as this is how I will understand it
I asked my professor for help and she said to start with X=Ak for some vale of k and to go from there.
Start by showing that $\forall A_k \ne \varnothing $, which is true because $2^k\cdot 1 \in A_k$.
Then show that $A_k \cap A_t = \varnothing$ for $k \ne t$, which again is true, otherwise if we suppose $\exists m \in A_k \cap A_t$ then $m=2^k\cdot m_k$ and $m=2^t\cdot m_t$. Given $k \ne t$ then $2^k \ne 2^t$ and we obtain that one of the $m_k$ or $m_t$ is even, which is a contradiction.
Then, the remaining part is to show that $\forall n \in \mathbb{N}, \exists k: n \in A_k$. If $n$ is odd then $n \in A_0$. If $n$ is even, then the fundamental theorem of arithmetic says $n=2^{k}\cdot p_{i_1}^{\alpha_1}\cdot p_{i_2}^{\alpha_2}...\cdot p_{i_l}^{\alpha_l}$, $p_{i_j}$-odd prime $\forall j=\overline{1,l}$, thus $n \in A_{k}$.