Computing the difference using the binary representations, taking into account that the terms disagree in the first spot and agree in the second, and finally applying the reverse triangle inequality, we get \begin{align*} |x-y| &= \left|\sum_{i=1}^{\infty} \frac{b_i-b_i'}{2^n}\right| = \left|\pm\frac{1}{2} + \sum_{i=3}^{\infty} \frac{b_i-b_i'}{2^n}\right| \\ &\geq \left|\left|\pm\frac{1}{2}\right|-\left|\sum_{i=3}^{\infty} \frac{b_i-b_i'}{2^n}\right|\right| \geq \left|\frac{1}{2} - \frac{1}{4}\right| = \frac{1}{4}. \end{align*}

Choose \(n\in\mathbb N\) large enough so that \(1/2^n < \varepsilon\text{.}\) Now suppose that \(x\in H\) has binary expansion

Define \(y\in H\) so that

That is, the bits of \(y\) agree with those of \(x\) in the first \(n\) spots, disagree with \(x\) in the \((n+1)^{\text{st}}\) spot, and finally agree with \(x\) again in the \((n+2)^{\text{nd}}\) spot.

Then, the numbers \(d^n(x)\) and \(d^n(y)\) satisfy the hypotheses of lemma Lemma 3, thus \(|d^n(x)-d^n(y)| \geq 1/4\text{.}\)

Let \(x\in I\) and choose \(n\in\mathbb N\) large enough so that

Now suppose that

Then,

is a periodic point in \(I\text{.}\)

We'll define \(x\) by specifying its binary expansion. We begin by writing down all possible finite binary strings:

We then concatenate these to obtain the binary representation of \(x\)

Now, let \(I\subset H\) be an open interval. We claim that there is some iterate of \(x\) in \(I\text{.}\) To see that, let \(L\) denote the length of \(I\) and choose \(n\in\mathbb N\) large enough so that

Let \(i\) be the smallest integer such that \(i/2^n \in I\text{.}\) Note that we must also have \((i+1)/2^n \in I\text{.}\) Thus, the dyadic interval \([i/2^n,(i+1)/2^n)\) is wholly contained in \(I\) and the first \(n\) bits of every point in that interval agree with \(i/2^n\text{.}\) So, let

and note that, by construction, the string \(b_1 b_2 \cdots b_n\) appears somewhere in the binary expansion of \(x\text{.}\) Thus, we can apply the doubling function to the point \(x\) some number, say \(m\text{,}\) times to obtain

The number \(d^m(x)\) is then an iterate of \(x\) that lies in \(I\text{.}\)