Exercise A0.III.6.07

Proposition

$\mathbb{Z}^{\mathbb{N}}$ is not isomorphic to $\mathbb{Z}^{\oplus \mathbb{N}}.$

Proof

Take any irrational real number $r$ and write down its continued fraction $[r_0; r_1, \ldots].$ Define the corresponding element of $\mathbb{Z}^\mathbb{N}$ as $(r_0, r_1, \ldots)$. This mapping is injective: two distinct real numbers have to differ at some point in their continued fraction expansion and their images in $\mathbb{Z}^\mathbb{N}$ will differ at the same position. So $\mathbb{Z}^{\mathbb{N}}$ is uncountable.

Now take any element $a$ of $\mathbb{Z}^{\oplus \mathbb{N}}.$ We can turn it into polynomial by taking

$$p_a = \sum_{k=0}a_kx^k.$$

Note that since $a$ has only finitely many non-zero components, this is indeed a polynomial and not a formal power series. A finite polynomial can be written down using some finite alphabet (why not 8-bit ASCII). We can then convert the code points into binary, concatenating them as 8-bit blocks together into a single binary number $n_a.$ This is an integer.

The process we used to encode $a$ is injective: distinct elements of $\mathbb{Z}^{\oplus \mathbb{N}}$ have to differ on some component, making their polynomial representations differ on the corresponding coefficient. This in turn results in differing bits, producing distinct integers.

Thus, $\mathbb{Z}^{\oplus \mathbb{N}}$ is countable.

There is no bijection between a countable and an uncountable set.

$\blacksquare$