A question about metric spaces that are compact and denumerably infinite Let S be any compact and denumerably infinite metric space and let d be the metric of S. We shall say that S satisfies condition C, if there exists at least one infinite sequence-------------p(1),p(2),.....,p(n),....of points of S such that (1) each point of S occurs once and only once in the sequence (2) the series d(p(1),p(2))+d(p(2),p(3))+.....+d(p(n),p(n+1))+....is convergent. If S satisfies condition C, an infinite version of the travelling salesman problem can be posed for the points of S. S always contains at least one limit point, since otherwise it could be covered by an infinite set of pairwise disjoint open balls, each containing a point of S-an impossibility if S is compact. It is easy to show that S cannot satisfy condition C if it contains more than one limit point. My question is: If S contains one and only one limit point, does S then necessarily always satisfy condition C?
 A: Let the space consist of a point $P$ and infinitely many other points $Q_n$ for $n\in\mathbb N$.  Let the distance from $P$ to $Q_n$ be $1/n$.  Let the distance from $Q_n$ to $Q_m$ for $m\neq n$ be $(1/n)+(1/m)$.  This seems to be a counterexample for your question.
A: The countable compact subspace $\ S:=\{q_0\ q_1\ \ldots\}\subseteq\ell^2\ $ of Hilbert space $\ \ell^2,\ $ given below, fails to have property C, i.e. it cannot be travelled along any finite length path (condition C was defined in the Question above):


*

*$q_0 := \mathbb 0\in \ell^2$

*$q_n\ $ has all coordinates $\ 0\ $ except for the $n$-th coordinate equal $\ \frac 1n\ $ for every $\ n=1\ 2\ \ldots$.


Now let $\ (p_1\ p_2\ \ldots)\ $ be a sequence of points of $\ S,\ $ in which each point $\ q_k\ $appears exactly one time. We will see that this routing path has infinite length:
$$\sum_{k=2}^\infty ||p_k-p_{k-1}||\ =\ \infty$$
PROOF
For each natural $\ M\ $ there exists natural $\ n\ $ such that
$$\{p_1\ \ldots\ p_n\}\ \supseteq\ \{q_1\ \ldots\ q_M\}$$
Let's use the following inequality:
$$\forall_{a\ b\in\ell^2}\quad \left(\ a\cdot b = 0\quad\Rightarrow\quad ||a-b||\ \ge\ \frac{||a||}{\surd 2}\ +\ \frac{||b||}{\surd 2}\ \right)$$
Thus:
$$\sum_{k=2}^n ||p_k-p_{k-1}||\ \ge\ \frac 1{\surd 2}\cdot\sum_{k=2}^n \left(||p_{k-1}||+||p_k||\right)\ =\ \surd 2\cdot\sum_{k=1}^n ||p_k||\ -\ \frac 1{\surd 2}\cdot\left(||p_1||+||p_n||\right)$$
hence
$$\sum_{k=2}^n ||p_k-p_{k-1}||\ \ge\ \surd 2\cdot\sum_{k=1}^M ||q_k||\ -\ \frac 1{\surd 2}\cdot\left(\frac 11+\frac 12\right)  $$
and finally:
$$\sum_{k=2}^n ||p_k-p_{k-1}||\ \ \ge\ \ \surd 2\cdot\log(M+1)\ -\ \frac{3\cdot\surd 2}4\quad\longrightarrow\quad\infty$$
for $\ M\rightarrow\infty\ $ (where $\ n\ $ depends on $\ M$)   END of PROOF


REMARK   @Andreas has provided an example which cannot be embedded isometrically into any Hilbert space (but I would never dream to call it exotic :-). Thus I decided to provide one in the Hilbert space $\ \ell^2$. Now harder work can start.

A: I'll show a counterexample in Euclidean $\ (\mathbb R^2\ \,d)$:

THEOREM There exists a countable compact $\ X\subseteq\mathbb R^2,\ $ which has exactly one limit point (hence $\ X\ $ is infinite), and such that every route $\ (p(1)\ p(2)\ \ldots)\ $ of $\ X\ $ has infinite length.

Here we understand that $\ (p(1)\ p(2)\ \ldots)\ $ being a route of $\ X\ $ means that all points $\ p(t)\ $ are different, and
$$ \{p(1)\ p(2)\ \ldots\}\ =\ X$$
and the length of such route is $\ \sum_{k=1}^\infty d(p(k)\ p(k+1))$.
PROOF   Let $\ O:=(0\ 0)\in\mathbb R^2.\ $ Let variables $\ k\ n\ $ run only through natural numbers $\ 1\ 2\ \ldots\ .\ $ Define
$$ X\ :=\ \{O\}\ \cup\ \left\{(\frac 1k\ \frac 1n)\ :\ \max(k\ n)\le 2\cdot\min(k\ n)\right\} $$
Consider $\ \ \delta(x) := \min_{y\in X\setminus\{x\}} d(x\ y)\ \ $ for every $\ x\in\mathbb R^2\setminus\{O\}.\ $ This minimum is well defined and satisfies:
$$ \delta\left(\left(\frac 1k \frac 1n\right)\right)\ \ge\ \frac 1{\max(k\ n)\cdot(\max(k\ n)+1)}$$
There are exactly $\ 2\cdot n+1\ $ points $\ \left(\frac 1r\ \frac 1s\right)\in X\ $ such that $\ \min(r\ s) = n.\ $ For each $\ 2\cdot n+1\ $ of such points $\ x\ $ we have $\ \delta(x)\ge \frac 1{2\cdot n\cdot(2\cdot n+1)}.\ $
Now consider an arbitrary route $\ (p(1)\ p(2)\ \ldots)\ $ of $\ X.\ $ For each of the $\ 2\cdot n+1\ $ points $\ p(t)\ $ such that it is one of points $\ x\ $ from above, we get:
$$ d(p(t)\ p(t+1))\ \ge\ \frac 1{2\cdot n\cdot(2\cdot n+1)}$$
Thus the sum of these $\ 2\cdot n + 1\ $ distances is at least $\ \frac 1{2\cdot n}.\ $ Thus (with a harmless abuse of notation):
$$\sum_{t=1}^\infty d(p(t)\ p(p(t+1))\ \gt \sum_{n=1}^\infty \frac 1{2\cdot n}\ \ge\ \infty$$
END of PROOF
