We verify each of the three conditions using the binary shift representation. **Sensitive dependence.** Fix $\delta = \frac{1}{2}$. Let $x \in [0,1)$ and $\varepsilon > 0$ be given. Choose $n$ large enough so that $2^{-(n+1)} < \varepsilon$. Let $y$ be the point obtained from $x = 0.a_1 a_2 \ldots$ by flipping the $(n+1)$-th binary digit (replacing $a_{n+1}$ by $1 - a_{n+1}$, with all other digits unchanged). Then \begin{align*} |y - x| = 2^{-(n+1)} < \varepsilon, \end{align*} while after $n$ applications of $F$, both $F^n(x) = 0.a_{n+1} a_{n+2} \ldots$ and $F^n(y)$ agree on all digits from position 2 onwards but differ in the first binary place, giving \begin{align*} |F^n(y) - F^n(x)| = \frac{1}{2} = \delta. \end{align*} **Topological transitivity.** Let $U, V \subset [0,1)$ be non-empty open sets. Pick $u = 0.a_1 a_2 \ldots \in U$ and $v = 0.b_1 b_2 \ldots \in V$. Since $U$ is open, there exists $N$ large enough that the point $u_N = 0.a_1 a_2 \ldots a_N b_1 b_2 \ldots$ (formed by concatenating the first $N$ binary digits of $u$ with all binary digits of $v$) still lies in $U$. Then $F^N(u_N) = 0.b_1 b_2 \ldots = v \in V$, so $F^N(U) \cap V \neq \varnothing$. **Density of periodic points.** Let $w = 0.a_1 a_2 \ldots a_m \ldots \in [0,1)$ and $\varepsilon > 0$ be given. For $N$ large enough that $2^{-N} < \varepsilon$, define $w_N = 0.\overline{a_1 a_2 \ldots a_N}$ (the periodic binary expansion obtained by repeating the block $a_1 \ldots a_N$). Then $|w_N - w| \leq 2^{-N} < \varepsilon$ and $F^N(w_N) = w_N$, so $w_N$ is a periodic point of period dividing $N$.