How to prove that $\{0,1\}^{\omega}$ is homeomorphic to $\{0,1\}^{\omega}$ $\times$ $\{0,1\}^{\omega}$?

Question

This is from a topology book that I'm using, how do I go about proving this? What map can serve as the homeomorphism?

Answer 1

Consider the map $f:\{0,1\}^{\omega}\to \{0,1\}^{\omega}\times \{0,1\}^{\omega}$ defined by $$f:\{x_n\}\mapsto (\{x_{2n-1}\},\{x_{2n}\})$$

Answer 2

Let's look at a countable product of spaces: $$Y=X_1\times X_2\times X_3\times\cdots.$$ Then $Y$ certainly is in bijection with $Y_1\times Y_2$ where $$Y_1=X_1\times X_3\times X_5\cdots$$ and $$Y_2=X_2\times X_4\times X_6\cdots.$$ Surely this bijection should be a homeomorphism? If so, if all the $X_i=X$, then $Y=X^\omega$ will be homoemorphic to $Y_1\times Y_2=X^\omega\times X^\omega$.

Answer 3

If you see the points as functions $f: \omega \to \{0,1\}$, one homeomorphism can be described as $F(f)=(f\restriction_O, f\restriction_E)$, where $O,E$ are the odd and even integers in $\omega$.

It is clearly a bijection, as $E \cup O = \omega$, etc. and it's continuous as $\pi_1 \circ F$ and $\pi_2 \circ F$ are essentially restriction maps (so "generalised projections", hence open), and similar remarks can be made on all $\pi_n \circ F^{-1}$, etc.