Is $\mathbb R^{\omega}$ homeomorphic to $\mathbb R^{\omega} \times \mathbb R^{\omega}$? As a study exercise, I'm trying to find a topological space $X$ which is homeomorphic to $X \times X$. I began thinking of simple examples involving $\mathbb R$ but then realized my best bet would be infinite dimensional, otherwise most of the time I couldn't have even a bijection. I'm almost sure that I've constructed a homeomorphism between $\mathbb R^{\omega} = \prod_{i\in\mathbb N} \mathbb R$ and the cartesian product with itself: $$ f:\mathbb R^{\omega} \times \mathbb R^{\omega}\to\mathbb R^{\omega} \atop{\left((a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\right)\mapsto (c_n)_{n\in\mathbb N}, c_n=\left\{\begin{align} &a_n, \quad n\text{ even} \\ &b_n, \quad n\text{ odd} \end{align}\right.}$$
Here we can see $f^{-1}((c_n))=((c_{2k}),(c_{2k+1}))$. Here the topologies involved are the usual ones, that is, the product topology in $\mathbb R^{\omega}$ (taking in $\mathbb R$ the standard metric topology), and the product topology again for the cartesian product with itself. I've tested for continuity and I'm pretty sure it's all good, but could someone tell me if I'm correct here? Thanks.
This happens to be the function I was able to construct, but my original intuition is that one sequence and two sequences next to each other are basically the same thing. Also if anyone happens to know a simpler example of the problem in my first sentence, please share.
P.S. I should be using simple topological arguments here; open and closed sets, antiimages, the likes.
 A: First, your formula
$$\left((a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\right)\mapsto (c_n)_{n\in\mathbb N}, c_n=\left\{\begin{align} &a_n, \quad n\text{ even} \\ &b_n, \quad n\text{ odd} \end{align}\right.$$
isn't quite correct, as is, you're dropping terms. You mean
$$\left((a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\right)\mapsto (c_n)_{n\in\mathbb N}, c_n=\left\{\begin{align} &a_{n/2}, \quad n\text{ even} \\ &b_{(n-1)/2}, \quad n\text{ odd} \end{align}\right.$$
judging by your formula for $f^{-1}$.
Yes, $f$ is a homeomorphism. To see that, use that a function into a topological product is continuous if and only if the function composed with any projection from the product to one of the factors is continuous.
You have
$$\pi_{2k}\circ f = \pi_k \circ \pi_a;\qquad \pi_{2k+1}\circ f = \pi_k \circ \pi_b,$$
and the projections $\pi_k$, $\pi_a$, and $\pi_b$ are continuous, hence $f$ is continuous. Conversely, you have
$$\pi_k \circ \pi_a \circ f^{-1} = \pi_{2k};\qquad \pi_k \circ \pi_b \circ f^{-1} = \pi_{2k+1},$$
which shows that $f^{-1}$ is also continuous.
A: It might be easier to argue using the discrete topology.  Similarly, you could use $\Bbb Z_2^\omega$ with the same argument as here
