Consider $S$ the set of all unit circles through the origin in $\mathbb{R}^3$. I'm trying to understand what this space looks like, assuming we use the obvious topology.
By 'obvious topology' I mean that circles that are close in $\mathbb{R}^3$ are also close in $S$. To formalize that, let's say that for two circles $c_1, c_2 \in S$ we define a metric $d_S(c_1, c_2)$ as e.g. $\max(\max_{p \in c_1} \min_{q \in c_2} d(p, q), \max_{q \in c_2} \min_{p \in c_1} d(q, p))$ with $d$ the standard euclidian metric (that is, find whichever point on $c_1$/$c_2$ is furthest from the other circle) and take the topology induced by $d_S$.
My question is: what does this topological space look like? It is clear to me that we can parametrize it from $S^2 \times S^1$ by first picking the center of our circle (this is the $S^2$) and then rotating it (this is the $S^1$). But it is not at all clear to me that the topological space ends up looking the same.
So what does this topological space look like?
Every circle is uniquely defined by plane in which it lies and line through its center, so your space $S$ is $\{ l, v \, | \, l \subset v \subset \Bbb R^3\}$, where $l$ is line and $v$ is plane. Most natural way to describe it would be factor $GL(3, \Bbb R)/UT(3, \Bbb R)$, where $UT$ is group of upper-triangular matrices with nonzeroes on diagonal (see flag variety). If you analyze this action and factor, you can conclude that it also can be described as factor of 3-dimensional sphere by free action of quaternion group embedded into it as subgroup of SU(2).