Suppose $\displaystyle A$ is denumerable. Prove that there is a partiion $\displaystyle P$ of $\displaystyle A$ such that $\displaystyle P$ is denumerable and for every $\displaystyle X \in P, X$ is denumerable.

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Note that for any partition $\displaystyle P$, it is not necessarily denumerable, and it is not necessarily that $\displaystyle X$ must be denumerable for all $\displaystyle X \in P$. For example, $\displaystyle A = \mathbb{Z}$ and $\displaystyle P = \{\mathbb{Z}^+, \{0\}, \mathbb{Z}^-\}$

I think I must construct some $\displaystyle P.$ Note that $\displaystyle A, P,$ and $\displaystyle X$ must be written as

$\displaystyle A = \{a_{11}, a_{12}, a_{13}, ..., a_{21}, a_{22}, a_{23}, ...,a_{31}, a_{32}, a_{33}...\}$ (A can be written like this since A is denumerable.)

$\displaystyle P = \{X_1, X_2, X_3,...\}$

$\displaystyle X_i = \{a_{i1}, a_{i2}, a_{i3},...\}.$

I cannot find any way to construct such $\displaystyle P$, other than what they should be by the characterization above.

Must I resort to proof by contradiction?

Thank you very much.

PS. Definition of partition, $\displaystyle 1. \bigcup P = A, 2. \forall X \in P, \forall Y \in P, X \cap Y = \emptyset, 3. \forall X \in P, X$ not equal $\displaystyle \emptyset$.