Prove that a denumerable set can be partitioned into two denumerable subsets I was wondering if this "proof" is sufficient in demonstrating a that a denumerable set $A$ can be partitioned into two denumerable subsets $A_1$ and $A_2$. Let $A$ be a denumerable set and define $A = A_1 \cup A_2$ where $A_1$ and $A_2$ are infinite subsets of $A$, and therefore they are also denumerable. Because $A$ is denumerable, a bijective function $f: \mathbb{N} \rightarrow A$ exists. Since $A = A_1 \cup A_2$ we can write $f:\mathbb{N} \rightarrow A_1 \cup A_2$. Because $f$ is bijective, the set $A_1 \cap A_2 = \varnothing$ (I think it might be sufficient to say that, because $f$ is a function, the two subsets must be disjoint). Finally, the denumaribility of the two subsets imply that they are not empty. So if we let the set $P = \\{A_1, A_2\\}$, the above information allows us to conclude that $P$ is a partition of $A$.

Your argument is circular: you’ve _started_ with denumerable sets that form a partition, when what you need to show is that such sets exist.

HINT: Use the fact that there’s a bijection $f:\Bbb N\to A$. Pick a partition of $\Bbb N$ into two denumerable sets; there’s one that’s especially obvious and easy to describe. If these sets are $N_0$ and $N_1$, consider the subsets $f[N_0]$ and $f[N_2]$ of $A$.