Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?

The answer is clearly yes for finite dimensional normed spaces: Define $T_\epsilon = \bigcap_{a \in A} \overline{B}(a, r + \epsilon)$. The $T_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.

For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x_n$ such that $x_n \in T_{\epsilon_n}$ with $\epsilon_n \to 0$, but I can't seem to make it work.

Note that an arbitrary choice of $x_n \in T_{\epsilon_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = \{ x : x_0 = 0, ||x|| \leq 1 \}$ then diam$(T_\epsilon) \geq 2$ because you can choose $c_0$ arbitrarily in $[-1, 1]$

Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center.

  • $\begingroup$ My first thought, given that completion is necessary, that compactness in norm topology suffices, and that this is a metric-looking condition, is: "Baire category"? $\endgroup$
    – Yemon Choi
    Apr 7, 2010 at 9:36
  • $\begingroup$ (Should explain that I am about to belatedly catch up on some sleep, so haven't given this the thought it deserves; but I just wanted to put up a first suggestion even if it turns out to be rubbish.) $\endgroup$
    – Yemon Choi
    Apr 7, 2010 at 9:38
  • $\begingroup$ I thought about it but couldn't seem to come up with a convincing way that it would work: There are no plausible looking open dense sets here and the $T_\epsilon$ are not meager (because $B(T_{\frac{1}{2} \epsilon}, \frac{1}{2} \epsilon) \subseteq T_{\epsilon}$ $\endgroup$ Apr 7, 2010 at 9:58

3 Answers 3


I believe that the property does not hold for all Banach spaces, but my counterexample is a little involved. If you've the patience then follow me through...

Let $V=\bigoplus_{n=1}^\infty \ell^n_2$ where $\ell^p_2$ is $\mathbb{R}^2$ with norm $\lVert\cdot\rVert_p$ (Note: $n$ is taking the role of $p$). For $i\geq1$ and $j\in\{0,1\}$ we have $e_{i,j}$, the $j^{th}$ standard basis vector of $\ell^i_2$ in $V$.

Give $V$ the norm $\lVert v\rVert=\sup_n\lVert v_n\rVert_n$.

Let $W=\{v\in V:\lVert v_n\rVert_n\to 0\}$. I assert that $W$ is a Banach space. Certainly every $e_{i,j}\in W$.

Let $A=\{e_{k,0}+e_{k,1}, e_{k,0}-e_{k,1}:k\geq 1\}$.

Fact: Let $r(A)$ be the infimum of radii of balls containing $A$. Then $r(A)\leq1$


Let $c_N=\sum_{i=1}^n e_{i,0}$. We wish to compute the distance of each point of $A$ from $c_N$.

For $k\leq N$ we have $\lVert c_N-e_{k,0}-e_{k,1}\rVert$ $=\lVert\sum_{i=1\ (i\not=k)}^Ne_{i,0}-e_{k,1}\rVert$ $=\sup\{\lVert e_{i,0}\rVert_i:i\leq N,i\not=k\}\cup\{\lVert-e_{k,1}\rVert_k\}$ $=1$ and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.

For $k>N$ we have $\lVert c_N-e_{k,0}-e_{k,1}\rVert$ $=\max(\lVert c_N\rVert,\lVert e_{k,0}+e_{k,1}\rVert_k)$ $= \max(1,(1+1)^\frac{1}{k})$ $= 2^\frac{1}{k}$ $\leq 2^\frac{1}{N}$ and similarly for $\lVert c_N-e_{k,0}+e_{k,1}\rVert$.

Thus $A\subseteq \overline{B}(c_N,2^\frac{1}{N})$ and so $r(A)\leq2^\frac{1}{N}$. Letting $N\to\infty$ we have $r(A)\leq 1$.


Fact: $A$ is not contained in a ball of radius $1$.


Suppose $A\subseteq \overline{B}(c,1)$. Then in particular for every $n$ we have $\lVert c-e_{n,0}-e_{n,1}\rVert\leq 1$ and thus $\lVert c_n-e_{n,0}-e_{n,1}\rVert_n\leq 1$. Similarly $\lVert c_n-e_{n,0}+e_{n,1}\rVert_n\leq 1$.

Simple consideration of $\ell^n_2$ shows that this implies $c_n=e_{n,0}$. Thus $\lVert c_n\rVert=1\not\to0$ and $c\not\in W$, contradicting the assumption.


  • 1
    $\begingroup$ Nice, J, and quite simple. So although $c_0$ has the property, for each $\epsilon > 0$ there is a space which fails the property and is $1+\epsilon$-isomorphic to $c_0$. BTW: Where are you? I could not tell from your MO page. $\endgroup$ Apr 8, 2010 at 0:48
  • $\begingroup$ Yes, that's true. I think I have a feel for what's going on here — the failure is something to do with the 'spikiness' of intersections of balls — but I can't put a finger on a nicer way to express it than the original question. (I'm in the UK) $\endgroup$
    – J Bytheway
    Apr 8, 2010 at 7:07

I think that http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654848-2/S0002-9947-1982-0654848-2.pdf [together with its references] provides us with several counterexamples [as well as with some remarkable examples], in the infinite-dimensional framework.


The answer may be affirmative when $V$ is a reflexive Banach space. Each $T_\epsilon$ is a closed bounded convex set. If the intersection of the $T_\epsilon$ is nonempty then each element of this intersection is an admissible centre. I found a paper online

Sharma, B. K., Dewangan, C. L. Fixed point theorem in convex metric space, Zb. Rad. Prirod.-Mat. Fak. Ser. Mat. 25 (1995), no. 1, 9-18 http://www.emis.de/journals/NSJOM/framepaper.htm

which asserts that every weakly compact convex subset of a Banach space has the property that a chain of nonempty closed convex subsets has nonempty intersection. Alas, they don't give a proof or reference for this.

By a theorem of W. F. Eberlein,

Weak Compactness in Banach Spaces, Proc. Natl. Acad. Sci. USA 33 (1947), 51–53

the closed unit ball in a Banach space is weakly compact iff the Banach space is reflexive.

  • $\begingroup$ Interesting. Thanks. I had an inkling that weak compactness might come into it: If you define $r(A) = \inf\{ t : \exists v, A \subseteq B(v, t) \}$ then we've got the result that if $V$ has a weakly compact unit ball then $r(A) = \sup \{ r(F) : F \subseteq A, \textrm{ finite } \}$, and every finite F is contained in $\overline{B}(c, r(F))$ for some c, but I couldn't come up with a convincing way to use that. $\endgroup$ Apr 7, 2010 at 10:58
  • 1
    $\begingroup$ Oh, actually this property is "obvious", in that closed convex sets of Banach spaces are closed in the weak topology, so compact in the weak topology. So reflexivity is a sufficient condition. It's not neccessary though: $l^\infty$ has the property as well, because $x_n = \frac{1}{2}(\sup a_n + \inf a_n)$ forms a center for $A$. $\endgroup$ Apr 7, 2010 at 11:06
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    $\begingroup$ In isolation, your first sentence isn't quite right, I think? closed bounded subsets of dual Banach spaces are compact in their weak-star topologies, but not necessarily in their weak topologies. Of course in a reflexive Banach space the two coincide $\endgroup$
    – Yemon Choi
    Apr 7, 2010 at 18:31
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    $\begingroup$ If $V$ is a dual space then $T_\epsilon$ is weak* closed by the weak* lower semicontinuity of the norm, so the property holds. The property also passes to norm one complemented subspaces, so being isometric to a dual space is not a necessary condition. $\endgroup$ Apr 7, 2010 at 18:40
  • $\begingroup$ I guess $c_0$ has the property, which suggests to me that all Banach spaces have it. Assume $r=1$. If $b(n)=\sup_A x(n)>1$ let $y(n) = b(n)-1$. If $c(n) = \inf_A x(n)<-1$ let $y(n) = c(n)+1$. If neither occurs let $y(n) = 0$. Then $B(y,1)$ contains $A$. $\endgroup$ Apr 7, 2010 at 20:52

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