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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?

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up vote 2 down vote accepted

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.

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Thanks Andreas, that is really very nice. Any infinite sequence in which each point of your space occurs at least once generates a series that grows faster (term by term) than some permutation of the Harmonic Series. Your metric space is certainly exotic. If n is any integer greater than 1, the point Q(n) is always "collinear with" the points Q(1) and p while p always lies "between" Q(1) and Q(n) on this "straight line". Could your space be isometrically embedded in any finite dimensional Euclidean space or even Hilbert space? I wonder whether there are any less exotic counterexamples. –  Garabed Gulbenkian Sep 27 '13 at 17:23
    
@GarabedGulbenkian I think your comment about collinearity almost proves that the space I suggested can't be embedded in Hilbert space. The straight line through Q(1) and P would have to contain all of the Q(n)'s, and every two of these would have to lie on opposite sides of P. But that's impossible, as there are only two sides of P on a line and we need infinitely many sides to accommodate all the Q(n)'s. –  Andreas Blass Sep 28 '13 at 12:57
    
Yes, that shows that your metric space is truly exotic. I tried, with no success, to find, some subsets of the Euclidean plane which I could prove to be counterexamples. Nor was I able to prove that such planar counterexamples could not exist. But your comment is also a proof that no subsets of the Euclidean straight line can be counterexamples. –  Garabed Gulbenkian Sep 28 '13 at 18:20
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