8. dyadic rationals

problem

Let $ℚ₂$ be the set of all dyadic rationals, i.e., rational numbers whose denominator in lowest terms is a power of $2$. Prove that $(0,1)∖ℚ₂$ is homeomorphic to the set of all sequences $x∈𝒞$ with infinitely many $0$s and $1$s.

Let $X⊆𝒞$ be the set of strings with infinitely many $0$s and $1$s each. Define a map $ϕ\colon 𝒞→[0,1]$ as follows:

$$ϕ(x) = ∑_{i=0}^∞ \frac {x(i)} {2ⁱ⁺¹},$$

i.e., interpret $x$ as a binary-digit expansion in the reals.

We claim that $ϕ$ restricted to $X$ is a homeomorphism to $(0,1)∖ℚ₂$. To show this we need to check that:

• $ϕ$ is well-defined
• $ϕ$ is bijective
• $ϕ$ is continuous
• $ϕ⁻¹$ is continuous

claim

$ϕ(X)=(0,1)∖ℚ₂$, i.e. for all $x∈X$ we actually have $ϕ(x)∈(0,1)∖ℚ₂$.

proof

Fix $x∈X$. It is straightforward to see that $ϕ(x)∈(0,1)$. To see that $ϕ(x)∉ℚ₂$, suppose otherwise. Note that all elements of $ℚ₂$ have finite binary-digit expansions, so let $y∈\{0,1\}^*$ be a finite sequence with $ϕ(y)=ϕ(x)$ (treating $y$ as having infinite trailing zeros). But $x≠y$ because $x$ has infinitely many $1$s; this can only occur if $x$ ends in all $1$s (do algebra to see this), again a contradiction because $x$ has infinitely many $0$s.