The "Circle" is a Vector Space? Consider the set of angles $C = [0, \ 2\pi)$ and, for all $x,y \in C$, define the $sum$ operation as the sum modulo $[0, \ 2\pi)$.
The identity element of the addition is the angle $0$. The inverse element exists.
Associativity and Commutativity hold.
I am not sure about the operation of $scalar \  multiplication$ $\cdot$.
1) Can I define the scalar multiplication (with identity element $1$) as $a \cdot x \ \text{mod } [0, \ 2\pi)$ for all $a \in \mathbb{R}$ and $x \in C$?
If so, seems that $a ( b \cdot x)  = (a b) \cdot x $ for all $a,b \in \mathbb{R}$ and $x \in C$.
Also seems that distributivity holds.
2) Is $(C,+,\cdot)$ a vector space then?
3) Is $(C,+,\cdot)$ metric for some distance function $d$?
 A: It can't be a vector space because scalar multiplication doesn't behave as required. For instance, given your algebraic operations, we'd have
$$\frac{1}{2} \star (2 \star \pi) = \frac{1}{2} \star 0 = 0$$
but
$$\left(\frac{1}{2} \cdot 2\right) \star \pi = 1 \star \pi = \pi \ne 0$$
so scalar multiplication doesn't satisfy the necessary conditions.
To disambiguate notation I've used $\star$ for scalar multiplication, and $\cdot$ for multiplication in $\mathbb{R}$.
A: You can certainly define a 'scalar multiplication' as you have. But that doesn't make it a vector space. 
Why? Because 1) if it were a vector space, it would obviously be 1-dimensional because any element in $(0, 2\pi)$ is a generator under your scalar multiplication but 2) the additive group of the angles is not isomorphic to $\mathbb{R}$, because it contains an element of finite order. 
Being a metric space, on the other hand, has nothing to do with the operations, and any subspace of a metric space is obviously a metric space under the inherited metric.
A: 1) and 2)  $C$ is not a vector space over $\mathbb{R}$, since if $x\in C$ such that $x\neq 0$, then $1\cdot x = \frac{x}{2\pi} \cdot (\frac{2\pi}{x}\cdot x) = 0$.
3) $C$ inherits a metric from the euclidean plane when viewing $C$ as the unit circle.
