Closed circle as a metric space If I take the closed unit circle, $C$, can I create a metric space $(C,d)$, where $d$ is the Euclidean-metric?
The reason for asking is that my text book says that all metric spaces are open, so:
(i) Is my example impermissible as a metric space? or
(ii) For points around the edge of the circle, $a$, the open balls $B(a,\delta)$ aren't your typical open balls, but rather the bit of the balls that intersects with $C$?
 A: Yes, $(C,d)$ is a metric space. And, yes, for points in $C$, the open balls are intersection of usual open balls (in $\mathbb{R}^2$) with $C$.
A: Every metric space $X$ is both open and closed in itself; note that $\varnothing$ is by definition both closed and open and that $\varnothing = X^{c}$. So $C$ is open in itself but, when considered as a subset of $\mathbb{R}^{2}$, is closed in $\mathbb{R}^{2}$.
I assume by a circle you really mean a circle rather than a disk. Let $a \in C$ and $\delta > 0$. Note that $B(a,\delta) := \{ x \in \mathbb{R}^{2} \mid |x-a| < \delta \}$ is an open ball that looks like an open disk. The set $\{ x \in C \mid |x-a| < \delta \} = B(a,\delta) \cap C$ is an open ball in $C$ but looks not like a ball.
A: If $(X,d)$ is a metric space and $A\subseteq X$, then $(A,d')$ is a metric space, where $d'$ is just $d$ restricted to $A$; namely, $d':=d|_{A\times A}$. Usually $d$ will be written instead of $d'$, and we'll just understand by context that it's the restriction. Now $A$ is open in the $d'$ metric, regardless whether or not it's open in the $d$ metric.
A different point of view is your (ii). If $x\in A$, then the $d$'-ball centered at $x$ with radius $r>0$ is equal to the intersection of $A$ with the $d$-ball centered at $x$ with radius $r$. In symbols,
$$
B_{d'}(x,r) = A\cap B_d(x,r).
$$
There is a partial converse to this statement as well. If $B_d(x,r)$ is a $d$-ball with $x\in X$, then the intersection $A\cap B_d(x,r)$ is open in the $d'$ metric. Indeed, if
$a \in A\cap B_d(x,r)$, then setting $\delta:=r-d(x,a)$ implies
$B_{d'}(a,\delta)\subseteq A\cap B_d(x,r)$. To see this, suppose $y\in B_{d'}(a,\delta)$. Then $y\in A$ by the fact that $d'$ only acts on pairs in $A$, and $y\in B_{d}(x,r)$ because $d(y,x) \leq d(y,a)+d(a,x) < \delta+d(a,x)=r$.
A: The entire space of a metric space and the empty set are open and closed.
You can create the metric space $(C,d_C)$ where $d_C$ is the restriction of the Euclidean metric $d_2$ to the points of $C$ .
The open balls of this metric space will be the open balls in the general Euclidean space intersecting $C$.
$(C,d_C)$ is called a subspace of the original space under the topology induced by the metric(in this case by the Euclidean metric $d_2$).
Also a subset $A$ of $C$ is open in $C$ iff exists an open subset $B$ of the general Euclidean space such that $A=B \cap C$.
The same is true a for a subset of $C$ to be closed in $C$.
