What topology makes $\mathbb{R}/\mathbb{Z}$ homeomorphic to $\mathbb{S}^1$ The circle $\mathbb{S}^1$ is homeomorphic to  $\mathbb{R}/\mathbb{Z}:=\{x+\mathbb{Z}\ |\ x\in [0; 1)\}$, but this answer states the topology is not the subspace topology. I wonder if it's the topology generated by half-open intervals $[a, b)$?
Also, what does the topology of $\mathbb{S}^1$ "look like", without embedding $\mathbb{S}^1$ in $\mathbb{R}^2$ or $\mathbb{C}$, but intrinsically? Is it made up of open intervals? Can we even think of open intervals in $\mathbb{S}^1$ without resorting to an "ambient space"?
 A: First, it is important to note that $S^1$ is not a subspace on $\mathbb{R}$, and so, the subspace topology is irrelevant here. Rather, it is a quotient of $\mathbb{R}$.
As to what the topology looks like: locally, $S^1$ looks exactly like $\mathbb{R}$. This means that small intervals in the circle are homeomorphic to real intervals. The difference between the line and the circle is a global one. That is, it cannot be seen by small creatures living on a circle. (This is somewhat similar to the difference between $S^2$ and the plane. Recall that it took people many years before they realized we were all living on a sphere).
Edit: In the answer mentioned in the post, the circle is identified with the interval $[0,1)$. While this identification may help to understand $S^1$ as a set or as a group (in the same way one can identify the ring $\mathbb{Z}/m\mathbb{Z}$ with the set $\{0,\ldots,m-1\}$), I find it very confusing in the topological aspect. If I had to describe the topology of the circle by means of subsets of $\mathbb{R}$, I would take the closed interval $[0,1]$ and glue both its endpoints to one another.
