Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

  1. which metric spaces are central?
  2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

A partial positive answer, when $\ S_1(\{a\ b\})\ $ always consists of a single point, was obvious from the moment the strongly convex spaces were introduced by Karol Borsuk:

    Operation $\ s\ $ is continuous for every compact strongly convex space.

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

A more specific fixed point theorem is a corollary of the above theorem, together with the easier part (2) of the theorem below:


  1. Every injective (i.e. hyperconvex) metric space is absolutely central.
  2. The central point operation $\ s\ $, defined for all non-empty bounded subsets, is continuous.

Corollary   Let $\ f:X\rightarrow X\ $ be an isometry of an arbitrary injective metric space such that $\ f(A)=A\ $ for a certain non-empty bounded $\ A\subseteq X.\ $ Then there exists $\ p\in X\ $ such that $\ f(x)=x$.

(I'd be willing to provide more material).

  • $\begingroup$ What is the difference between absolutely central and strongly central as defined here? (Maybe the latter refers to non-empty finite subsets?) $\endgroup$ – Benjamin Dickman Dec 24 '14 at 7:23
  • $\begingroup$ It was indeed a typo. I missed totally in totally bounded. @BenjaminDickman -- your proposition is valuable. I don't believe it that I missed it for all these years; but yes, your variant is new to me! :-) $\endgroup$ – Włodzimierz Holsztyński Dec 24 '14 at 7:30
  • $\begingroup$ REFERENCE: -- S.Mazur and S.Ulam, Sur les transformations isometriques d`espaces vectoriels normes, C.R. Acad. Sci., Paris, 194 (1932), pp. 946-948. Hmm, 82 years ago. (I'd appreciate to get this reference MO-professionally :-) into the text of the Question; I'll then learn and will do it myself on eventual next occasions). $\endgroup$ – Włodzimierz Holsztyński Dec 24 '14 at 7:42

No, it's not:

THEOREM (Example)   There exists a compact central metric space for which the central point operation is not continuous.

PROOF (Construction)   Consider the spherical distance. For any two points that are not antipodals, $S_1$ is a singleton. For antipodal points, like the North and South poles, you get their equator.

Our metric space will be a part of the sphere that contains the North and South poles, but no other antipodal points and contains the shortest path between any two of its points (except the North and South poles). It will be the set of points whose longitude is between, say, 0 and 20 degrees (including the North and South poles). Then for any two points other than the North and South poles $s=S_1$, while for them $s=S_2$ is the point of the equator with 10 degrees longitude. This shows that $s$ is not continuous as we can keep the South pole fixed and converge to the North pole along the 0 longitude line.

  • $\begingroup$ Thank you, @domotorp -- your solution is so nice and so simple! $\endgroup$ – Włodzimierz Holsztyński Dec 24 '14 at 21:32
  • $\begingroup$ Your construction reminds me somewhat of my own--by necessity much more involved--different theorem/example from years ago, about the non-continuity of $\ s\ $ for a complete separable and contractible strongly convex metric space (Karol Borsuk's definition/notion). $\endgroup$ – Włodzimierz Holsztyński Dec 24 '14 at 21:38

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