The generated normalized Fibonacci lifts are dense in SU(2)

The matrices $B_j$ are unitary by [F1], and $z$ has modulus one. Their determinants satisfy $\det (z{B}_j)=z^2\det (B_j)=e^{3\pi i\mathrm{/}5}{e}^{-3\pi i\mathrm{/}5}=1$ by [F1]–[F2], so each lifted generator lies in $SU(2)$. Let $q:SU(2)\to PU(2)$ be the quotient map with kernel $\{I,-I\}$. Multiplication by the scalar $z$ does not change a projective class, so $q(\mathrm{\Lambda })=\mathrm{\Gamma }$. By [A1] and [F3], $\mathrm{\Gamma }$ is dense, so [F4] gives density of $q^{-1}(\mathrm{\Gamma })$. Every fiber of $q$ has exactly two elements, which yields $q^{-1}(\mathrm{\Gamma })=\mathrm{\Lambda }\cup (-\mathrm{\Lambda })$.

Let $K=\bar{\mathrm{\Lambda }}$ in $SU(2)$. It is a subgroup: for $x,y\in K$ and any neighborhood $O$ of $x{y}^{-1}$, continuity of $(u,v)\mapsto u{v}^{-1}$ gives neighborhoods $U,V$ of $x,y$ with $U{V}^{-1}\subseteq O$; since $x,y$ lie in the closure, choose $u\in U\cap \mathrm{\Lambda }$ and $v\in V\cap \mathrm{\Lambda }$, so $u{v}^{-1}\in O\cap \mathrm{\Lambda }$. Hence $x{y}^{-1}\in K$. By step 1.1, $q^{-1}(\mathrm{\Gamma })=\mathrm{\Lambda }\cup (-\mathrm{\Lambda })$ is dense; taking closures of this finite union gives $K\cup (-K)=SU(2)$. If $-I\in K$, then $-K=K$ and the cover forces $K=SU(2)$. If $-I\mathrm{\notin }K$, then $K\cap (-K)=\mathrm{\varnothing }$: an intersection would give $x=-y$ for $x,y\in K$, hence $-I=y^{-1}x\in K$. Both sets are closed, so each is also open as the complement of the other; this separates the connected group $SU(2)$ into two nonempty clopen sets, contradicting [F5]. Therefore $-I\in K$ and $K=SU(2)$. Only [A1] is used as a choice axiom.