Let us consider the group of continuous functions $C(S^1, S^1)$ from the circle to itself with the compact open topology. Does it have a chance to contain a dense cyclic subgroup? 


Unfortunately it is not compact so we cannot apply the standard argument about compact, connected groups.



When are function groups monothetic?
No. Actually, for every, say metrizable space $X$ that is not totally disconnected, $C(X,S^1)$ is not even monothetic for the pointwise convergence topology (that induced by inclusion in $(S^1)^X$ with the Tychonov topology). Indeed, let $C$ be a connected component of $X$ not reduced to a point.


If $f$ is constant on $C$, then so is every pointwise limit point of $\mathbf{Z}f$ (I'm writing $S^1=\mathbf{R}/\mathbf{Z}$ additively), so $\mathbf{Z}f$ is not dense. If $f$ is not constant on $C$, then there exists $x\in C$ such that $nf(x)=0$, and this property is shared by every pointwise limit point of $\mathbf{Z}f$. In both cases we conclude that $\mathbf{Z}f$ is not dense.




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However, for $X$ Cantor (or more generally compact metrizable totally disconnected), I claim that $C(X,S^1)$ is monothetic, by showing that the set $W$ of $f$ such that $\mathbf{Z}f$ is dense, is $G\_\delta$-dense.


Indeed, there is a countable basis of the topology $\mathcal{D}$, such that every $U\in\mathcal{D}$ is given as follows: a number $d=d(U)$, a clopen partition $X=K\_1\sqcup \dots K\_d$, open intervals $I\_1,\dots,I\_d$, such that $U$ is the set of $f$ such that $f(K\_j)\subset I\_j$ for all $j$.


For $U\in\mathcal{D}$, define $U^\sharp=\{f:\exists n\in\mathbf{Z}:nf\in U\}$: as a union of open subsets, $U^\sharp$ is open.


Then $W=\bigcap\_{U\in\mathcal{D}}U^\sharp$ is $G\_\delta$. Hence, to show that it is dense, it is enough to show that $U^\sharp$ is open for every $U$. Indeed, fix $U$ as above, $\varepsilon$ and $f$. Then there exists a clopen partition $(L\_k)$, such that each $L\_k$ is contained in some $K\_j$, and a function $g$ constant (say equal to $g\_j$) on each $L\_j$, such that $\|f-g\|\_\infty\le\varepsilon$. We can suppose that the $g\_j\in\mathbf{R}/\mathbf{Z}$ form a $\mathbf{Z}$-free family. Then a density argument in finite-dimensional tori shows that $ng$ belongs to $U$ for some $u$, that is, $g\in U^\sharp$. This shows density.


