The Baire space is homogeneous,

Question

it's kwown that the Cantor space $2^{\omega}$ is a homogeneous topological space (because it is a topological group). Does anyone have any idea why $\omega^{\omega}$ is a homogeneous topological space? Thanks.

Answer 1

If $x,y\in\omega^\omega$, for each $n\in\omega$ let $f_n:\omega\to\omega$ be the map the interchanges $x_n$ and $y_n$ and leaves everything else fixed. Let

$$h:\omega^\omega\to\omega^\omega:z\mapsto\langle f_n(z_n):n\in\omega\rangle\,;$$

then $h(x)=y$, and $h$ is an autohomeomorphism of $\omega^\omega$.

Another approach is to note that $\omega^\omega$ is homeomorphic to $\Bbb Z^\omega$, and that the product topology on $\Bbb Z^\omega$ is the same as the order topology induced by the lexicographic order $\preceq$. Now let $\sigma,\tau\in\mathbb{Z}^\omega$, and define $\delta\in\mathbb{Z}^\omega$ by $\delta(n)=\tau(n)-\sigma(n)$. Then the translation $$s:\mathbb{Z}^\omega\to\mathbb{Z}^\omega:\varphi\mapsto\varphi+\delta=\langle\varphi(n)+\delta(n):n\in\omega\rangle$$ is an order-automorphism of $\langle\mathbb{Z}^\omega,\preceq\rangle$ taking $\sigma$ to $\tau$ and thus also an autohomeomorphism of $\Bbb Z^\omega$ with the order topology induced by $\preceq$.

Answer 2

It's also a topological group, just view it as $\Bbb Z^\omega$ in the obvious coordinatewise group operation. $\Bbb Z \simeq \omega$ topologically after all.