With the product topology $(X^\omega)^\omega$ and $X^\omega$ are homeomorphic From some lecture notes:

Here $a_n$ are sequences. The standard bijection in question is $\langle n, m \rangle = 2^n(2m + 1) - 1$. The topology on $Y^\omega$ is the product topology, both for $Y = X$ and $Y = X^\omega$.
I don't see why $h^{-1}$ is continuous, as any small change in $a \in X^\omega$ according to the metric (which induces the product topology):
$$ d(a, b) = \begin{cases}
0 & \text{ if } a = b \\\\
2^{-n} & \text{ if } a \upharpoonright n = b \upharpoonright n \land a(n) \neq b(n)
\end{cases}$$
impacts the sequence $(h^{-1}(a))(0)$ already. So that no matter the distance $d_{X^\omega}(a, b)$, we have $d_{(X^\omega)^\omega}((h^{-1}(a)), (h^{-1}(b))) = 1$.
 A: It’s easier to work directly with the product topology.
${^\omega}({^\omega}X)$ has a base consisting of sets $B$ of the following form. Let $F\subseteq\omega$ be finite, and for each $n\in F$ let $F_n$ be a finite subset of $\omega$. For each $n\in F$ and $k\in F_n$ let $U(n,k)$ be an open set in $X$. Let
$$B_n=\left\{a\in{^\omega}X:a(k)\in U(n,k)\text{ for each }k\in F_n\right\}\,,$$
a basic open set in ${^\omega}X$, and let
$$B=\left\{\langle a_n:n\in\omega\rangle\in{^\omega}({^\omega}X):a_n\in B_n\text{ for each }n\in F\right\}\,.$$
Let $a\in{^\omega}X$. Then $a\in h[B]$ iff there is $\langle a_n:n\in\omega\rangle\in{^\omega}({^\omega}X)$ such that $a=h(\langle a_n:n\in\omega\rangle)$ and $a_n(k)\in U(n,k)$ for each $n\in F$ and $k\in F_n$. But if $a=h(\langle a_n:n\in\omega\rangle)$, then $a(\langle n,k\rangle)=a_n(k)$ for each $n\in F$ and $k\in F_n$, so $a\in h[B]$ iff $a(\langle n,k\rangle)\in U(n,k)$ for each $n\in F$ and $k\in F_n$. That is,
$$h[B]=\left\{a\in{^\omega}X:a(\langle n,k\rangle)\in U(n,k)\text{ for each }n\in F\text{ and }k\in F_n\right\}\,,$$
which is open in ${^\omega}X$. Thus, $h$ is a continuous, open bijection and is therefore a homeomorphism.
A: A function $f: Y \to {{}^\omega}X$ is continuous (where the power has the product topology) iff $\forall n \in \omega: \pi_n \circ h: Y \to X$ is continuous where $\pi_n: {{}^\omega}X \to X;\, \pi_n(a)  =a(n)$ for all $n$ (the projection maps, which are always continuous on products). This is a standard fact about product topologies.
So with $\mathbb{h}_X$ we have that for $k \in \omega$ that $\pi_k \circ \mathbb{h}_X = \pi_n \circ \pi_m$ where $n,m$ are the unique members of $\omega$ such that $\langle n,m \rangle = k$, and this is a composition of two continous projections, hence continuous. So $\mathbb{h}_X$ is continuous.
For the inverse map (call it $\mathbb{k}$, say) we use a similar argument,
but then we have a "power of powers" so we need that $\pi_n \circ \pi_m \circ \mathbb{k}$ is continous for all $n,m$ but this just equals $\pi_{\langle n,m\rangle}$ (or the other order, it's late) and so just a (continous) projection again.
So both $\mathbb{h}_X$ and its inverse are continuous.
