can anyone please guide me through b? I am confident in my proof for a, however, I would like some guidance for b. Does b just follow from a? The space we are dealing with is Hausdorff. Thank you!
Let $K=\{0,1,2,3\}$ and let $C=\{0,1\}$. Prove:
$(a)$ For any $n\in \mathbb{N}, K^n$ with the product topology is homeomorphic to $C^{2n}$ with the product topology.
$\textbf{Proof:}$ Since the intention is that $K$ and $C$ both have the discrete topology, then the product spaces $K^n$ and $C^{2n}$ are homeomorphic. The sets $K^n$ and $C^{2n}$ have both cardinality $2^{2n}$ and therefore there exists a bijection between them. The product topologies on $K^n$ and $C^{2n}$ are both discrete. More generally, for any finite, discrete topological space $X$ and any integer $k\ge 1$, the product topology on $X^k$ is discrete. Any bijection between discrete topological spaces is homeomorphic.
$(b)$ The spaces $K^{\mathbb{N}}$ and $C^{\mathbb{N}}$, both with the product topology, are homeomorphic to each other.
Best Answer
As to the first, the finite spaces are Hausdorff and thus must be discrete. The finite products are also discrete and have the same size, and so are homeomorphic. Done.
That the countable powers of finite discrete spaces are homeomorphic is a consequence of Brouwer’s theorem that all countable metrisable zero-dimensional spaces without isolated points are homeomorphic. But there is an easier argument in this case because you can write the power of $C$ as the power of $K$ using an odd/even partition of $\mathbb{N}$. Just group by 2, and use only that $C \times C \simeq K$.