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In a metric space, let's denote with $B(x,r)$ the open ball of center $x$ and radius $r$ and with $E(x,r)$ the closed ball. While waiting for an official term, let's call naïf a metric space such that $\overline{B(x,r)}=E(x,r)$ holds for all $r>0$ and $x\in X$.

I was answering this question: apparently, there is a theorem according to which in a compact naïf metric space (open) balls are always connected.

When it came to providing examples of non-compact naïf metric spaces with a disconnected open ball, all I could think of was $\Bbb Q$. However, I was wondering

Is there a naïf complete metric space with a disconnected open ball?

Sneaky bonus question: is there a naïf complete metric space with a disconnected closed ball? $\ddot\smile$

As a preliminary observation, by virtue of that theorem the counterexample-ball cannot be relatively compact. Therefore, closed subspaces of $\Bbb R^n$ with the induced metric won't make good counterexamples.

Looking for spaces where no closed ball is compact - say, an appropriate closed subset of $\ell^2$ or $L^\infty$ - seems the most natural choice, but nothing is occurring to me right now.


Warning: For the sake of clarity, the following conditions are deceivingly similar to naïveté on paper, but they are not the same thing:

  1. "Every open ball is closed": balls that are closed sets are not necessarily closed balls. In fact, the only naïf metric space where this condition holds is the singleton, because we'd need $B(x,r)=\overline{B(x,r)}=E(x,r)$, which always fails for $r=d(x,y)$.

  2. "The closure of any open ball is a closed ball": same center and radius are needed. $\Bbb N$ with the standard distance satisfies this, but it is not naïf because $\overline{B(x,n)}\ne E(x,n)$ for all positive integers $n$.

However: This criterion describes an equivalent condition to naïveté: "A metric space is naïf if and only if for all $r>0$ and for all $x\ne y\in X$ there is $z\in X$ such that $d(y,z)<\varepsilon$ and $d(x,y)>d(x,z)$".

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I think that the complete Erdös space $E_c$ is a candidate. This consists of all sequences $(x_n) \in \ell_2$ such that $x_n \in \{0\} \cup \{ \frac{1}{k}: k \in \mathbb{N}^+ \}$ for all $n$, in the subspace topology.

This is complete one-dimensional and totally disconnected. I think that is "naïf" too (please check).

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    $\begingroup$ Note: just being ultrametric doesn't guarantee that $\overline{B(x,r)} = E(x,r)$; for example, this isn't true in the $p$-adics. It might hold here, but it doesn't follow simply from the ultrametric property, so you might want to provide an argument for that statement. $\endgroup$ – Stahl Jan 29 '17 at 19:26
  • $\begingroup$ You're quite right, I misread the wikipedia entry. it holds here for all radii not equal to some $\frac{1}{2^k}$, though, not for all $r$, just for a base of balls. Open balls are all closed and closed balls are also open. $\endgroup$ – Henno Brandsma Jan 29 '17 at 20:09
  • $\begingroup$ @G.Sassatelli I changed the answer ,so please revisit $\endgroup$ – Henno Brandsma Jan 30 '17 at 18:17
  • $\begingroup$ @HennoBrandsma I saw it. I'm checking with this criterion. It is promising. $\endgroup$ – user228113 Jan 30 '17 at 19:32
  • $\begingroup$ @HennoBrandsma I fear it is not naïf, though: if you consider for the criterion I linked $\varepsilon=\frac16$ and the sequences $x=(1,0,0,\cdots)$ and $y=\left(\frac12,0,0,\cdots\right)$, you have that $\lVert z-y\rVert <\frac16\implies z_1=\frac12$. However, this means that $\lVert z-x\rVert\ge \frac12=\lVert x-y\rVert$. $\endgroup$ – user228113 Jan 30 '17 at 19:48

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