A free action is not necessarily a covering space action: counterexample In p.73 of Hatcher's algebraic topology, he gives an example of a free action which is not a covering space action.

An example is the action of $\mathbb{Z}$ on $S^1$ in which a generator of $\mathbb{Z}$ acts by rotation through an angle $\alpha$ that is an irrational multiple of $2\pi$. In this case each orbit $\mathbb{Z}y$ is dense in $S^1$, so condition ($\ast$) cannot hold since it implies that orbits are discrete spaces.

($\ast$) is the condition of a covering space action. I am confused about the meaning of "discrete spaces" and how it connects with denseness in this case. Can anyone work out this example explictly?
 A: Recall that condition $(*)$ is the following

A discrete space is just a space endowed with the discrete topology, that is the topology in which every subset is an open set. From the axiom of a topology, you get immediately that a topological space is discrete if and only if singletons are open sets.


*

*First we can prove that if $(*)$ holds for an arbitrary space $Y$, orbits are discrete spaces which means that the induced topology on the orbit space is the discrete topology. Let $x\in Y$ and let $H_x$ be the orbit space of $x$. Let $y\in H_x$ and $U$ as in $(*)$. Then any element $z$ in the orbit $H_x$ (which is also $H_y$) can be written $z=g(y)$ for some $g\in G$. Then if $z\in U$ we have 
$$z\in U\cap g(U)=id(U)\cap g(U)$$
so by assumption $g=id$ and then $z=y$. This shows that $H_x\cap U=\{y\}$ which means that $\{y\}$ is an open subset of $H_x$ (endowed with the induced topology). Hence $H_x$ is discrete.

*Now  being a discrete subspace and being dense are not contradictory assumptions: if $A$ is any discrete space, $A$ is a discrete subspace of $A$ and is dense is $A$. But in your case you have even better than denseness: if $y\in H_x$, every neighborhood of $y$ in $S^1$ intersects $H_x-\{y\}$ (you have a sequence in $H_x-\{y\}$ that converges to $y$). Therefore any neighborhood of $y$ in $H_x$ intersects $H_x-\{y\}$ ($y$ is a limit point). This is in contradiction with the fact that $H_x$ is discrete, so $(*)$ can't hold.
I hope this helps!
Edit: following your comment I will add some details regarding the last point. If $H$ denotes some orbit, $H$ has the property that $S^1-H$ is dense in $S^1$. This is pretty clear because $H$ is countable, and every non-empty open subset of $S^1$ is uncountable. Now we are left to prove the following:

If $X$ is a metric space and $H\subset X$ is dense and $X-H$ is dense, then every $h\in H$ is the limit of a sequence of elements of $H-\{h\}$.

We just have to prove that for every $h\in H$ and $\epsilon>0$ there exist $h^\prime\in H-\{h\}$ such that $d(h,h^\prime)<\epsilon$. Let $h\in H$ and $\epsilon >0$. Because $X-H$ is dense there exists $z\in X-H$ such that $d(h,z)<\epsilon/2$. Now because $H$ is dense there exist $h^\prime\in H$ such that $$d(z,h^\prime)<\min\{\epsilon/2,d(h,z)\}$$
Now from triangular inequality you have that $$ d(h,h^\prime)>d(h,z)-d(z,h^\prime)>0$$
so $h\neq h^\prime$ (i.e $h^\prime\in H-\{h\})$ and $$d(h,h^\prime)<d(h,z)+d(z,h^\prime)<\epsilon/2+\epsilon/2=\epsilon.$$
(just do a drawing to understand what is happening).
