Why is $2^{\mathbb{N}}$ Homeomorphic to $2^{\mathbb{N}^{< \mathbb{N}}}$?

Question

This question is based off a problem from Classical Descriptive Set Theory by Kechris. In this book, Kechris makes the claim that, when $\{0, 1\}$ is endowed with the discrete topology, then the Cantor space $\{0, 1\}^{\mathbb{N}}$ is homeomorphic to the set $\{0, 1\}^{\mathbb{N}^{<\mathbb{N}}}$ (both endowed with the product topology). I am struggling to see why this is true, and I cannot think of an explicit homeomorphism between the two spaces. Could someone explain why this is the case?

Answer 1

Let $I.J$ be infinite index sets. Then $\{0,1\}^I \simeq \{0,1\}^J$ (as spaces) iff the sets $I$ and $J$ have the same cardinality. A shuffle of coordinates is a homeomorphism (if $f: I \to J$ is a bijection, $\hat{f}: \{0,1\}^J \to \{0,1\}^I$ defined by $\hat{f}(g)=g \circ f$ is a homeomorphism)

For the reverse note that the weight of $\{0,1\}^I$ equals $|I|$ and weight ( the minimal size of a base) is a topological invariant.