Continuum-distanced complete, ultrametric space

Question

Our professor asked us to find a complete metric space where the intersection of nested closed balls can be empty.

The following space is such an example, and I would like to learn more on it (since my efforts to find more info on the internet have so far been unfruitful).

Consider $X$ the set of continuous, real-valued functions defined on $]0,\infty[$. Define $\|f\|=\sup\{x>0:f(x)\neq 0\}$, with $\|0\|=0$. Then $\|\cdot \|$ satisfies $$\|f+g\|\leq\max(\|f\|,\|g\|)$$ $$\|f\|=0\iff f=0$$ so that $d(f,g)=\|f-g\|$ defines an ultrametric distance over $X$.

$(X,d)$ is complete: if $(f_n)_{n\geq 1}$ is Cauchy, w.l.o.g. we may suppose $d(f_{n-1},f_{n})<2^{-n}$. Then $f_n,f_{n+1},f_{n+2},\ldots\}$ agree on $[2^{-n},\infty[$. Hence the sequence converges pointwise in the usual real topology and it is clear that the pointwise limit is continuous and also the limit in $(X,d)$ of $(f_n)$.

Let $g_n(x)=\min(n^2,1/(x-1)^2)$. Then, if $B_c$ means "closed ball", $\cap_{n\geq 1}B_c(g_n, 1+1/n)$ is nested and empty (it "must" contain $x\mapsto 1/(x-1)^2$ which is discontinuous).

I would like to know more about why this example appears. What does the community know about it? In particular, this question seem very natural to me:

• Is there a simple way to modify $\|\cdot\|$ so as to turn into an actual norm?

Answer 1

This is impossible, because

for any complete normed space $X$ and any nonincreasing sequence $(B_n)$ of nonempty closed balls in $X$ we have $\bigcap_n B_n\ne\emptyset$.

Indeed, take any nonincreasing sequence $(B_n)$ of nonempty closed balls in $X$. For each $n$, let $x_n$ and $r_n\ge0$ be, respectively, the center and radius of the ball $B_n$. The crucial observation is that for each $n$ $$\|x_{n+1}-x_n\|\le r_n-r_{n+1}; \tag{1}\label{1}$$ find the proof of \eqref{1} below. So, $r_n\downarrow r$ for some real $r\ge0$ and, by telescoping and the norm inequality, $\|x_m-x_n\|\le|r_n-r_m|\to0$ as $m,n\to\infty$. So, by the completeness of $X$, $$x_n\to x$$ for some $x\in X$.

Now note that for all $m\ge n$ we have $x_m\in B_m\subseteq B_n$ and hence $\|x_m-x_n\|\le r_n$. Letting $m\to\infty$, we get $\|x-x_n\|\le r_n$, that is, $x\in B_n$, for all $n$, so that $x\in\bigcap_n B_n$ and hence $\bigcap_n B_n\ne\emptyset$. $\quad\Box$

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Proof of \eqref{1}: Let $$y:=x_{n+1}+r_{n+1}\frac{x_{n+1}-x_n}{\|x_{n+1}-x_n\|}$$ if $x_{n+1}\ne x_n$, and $y:=x_{n+1}+r_{n+1}u$ for an arbitrarily chosen unit vector $u\in X$ if $x_{n+1}=x_n$. Then $y\in B_{n+1}\subseteq B_n$, so that $$r_n\ge\|y-x_n\|=\|x_{n+1}-x_n\|+r_{n+1},$$ which yileds \eqref{1}. $\quad\Box$

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Remark 1: It is easy to see that condition \eqref{1} is, not only necessary, but also sufficient for $B_{n+1}\subseteq B_n$.

Remark 2: It follows that (of course) the condition of the completeness of $X$ cannot be dropped. Indeed, let $X$ be the vector space of all real-valued sequences with only finitely many nonzero values, endowed with (say) the $\ell^\infty$ norm. Let $(r_n)_{n\ge0}$ be any strictly decreasing sequence of real numbers converging to $0$. For each $n\ge1$, let $a_n:=r_{n-1}-r_n>0$ and, for each $n\ge0$, let $x_n:=(a_1,\dots,a_n,0,0,\dots)$. As before, let $B_n$ be the closed ball in $X$ of radius $r_n$ centered at $x_n$. Then, by Remark 1, $B_n\subseteq B_{n-1}$ for each $n\ge1$. Suppose now that $B:=\bigcap_n B_n\ne\emptyset$, so that for some $y=(y^{(1)},y^{(2)},\dots)$ we have $y\in B$. Then for each natural $k$ and $n$ such that $n\ge k$ we have $|y^{(k)}-a_k|\le r_n$. Letting now $n\to\infty$ and recalling that $r_n\downarrow0$, we see that $y^{(k)}=a_k\ne0$ for all $k$, which contradicts the condition that $y\in B\subseteq X$.

Answer 2

Let's put put the question in a bit more general context: Which type of metric spaces admit the property that nested closed balls can have empty intersection:

(1) Complete: Yes (as in the original post)

(2) Complete, normed: No (as in the answer)

(3) Compact: No: Take a converging subsequence of the midpoints of the balls

(4) Separable, complete, discrete: Yes, see below

(5) Convex closed subset of a separable Banach space: Yes, see below

(6) Wasserstein space of some Polish metric space: ???

...

Regarding (4): One can simplify the original example by only considering the sequence of midpoints from the original post. This should give something like

$(X = \mathbb N,d)$ with $d(m,n) := r(m\wedge n)$ where $\wedge$ denotes the Minimum and $r(n) := 1+ 1/n$. Then, $$ B_{1+1/n}(n) = \{n,n+1,\ldots\} $$ Clearly, these balls are nested and the intersection is empty, and the space is separable and discrete.

Regarding (5): we can isometrically embed the space $(X,d)$ into $\ell_\infty(\mathbb N)$ via a map $\phi : \mathbb N \to \ell_\infty(\mathbb N)$, $$ n \mapsto r_n 1_{\cdot \geq n}. $$ (Note that this is not the standard-embedding via the distance). Let $Y \subset \ell_\infty(\mathbb N)$ be the closure of the convex hull of the image of $\phi$.

Now I claim that within $Y$, the balls $B_{r(n)}(\phi(n))$ are nested with empty intersection.

We first show $B_{r(n)}(\phi(n)) \supset B_{r(n+1)}(\phi(n+1))$ in $Y$. For non-negative $f \in \ell_\infty(\mathbb N)$, we have $$ f \in B_{r(n)}(\phi(n)) \Leftrightarrow f\leq r(n) + \phi(n) $$ and the inclusion of balls follows as $r(n) + \phi(n) \geq r(n+1) + \phi(n+1)$ and as $Y$ only contains non-negative functions.

We now show empty intersection (clearly the balls intersect in $\ell_\infty(\mathbb N)$, but not in $Y$).

We can see that $$ \mathbb R_+^{\mathbb N} \cap \bigcap_k B_{r(n)}(\phi(n)) = \{f \in \ell_\infty(\mathbb N) : 0 \leq f \leq 1\}. $$ So we are left to show that no function in $Y$ is upper bounded by one. Suppose such a function exists, call it $f_0$.

Clearly, $f_0\geq 0$ and $f_0$ is increasing and $ \|f_0\|_\infty = \lim_{k\to \infty} f_0(k) \geq 1$ (as $f_0 \in Y$). We will refine this to show that $\|f_0\|_\infty>1$.

Let $k \in \mathbb N$ be fixed with $f_0(k)>0$. For $\varepsilon >0$ let $f_\varepsilon$ in the convex hull of $\phi(\mathbb N)$ such that $\|f_0 - f_\varepsilon\|_\infty < \varepsilon$.

Choosing $\varepsilon$ small, we can ensure $f_\varepsilon(k) \geq 2(k+1) \varepsilon$. We can write $f_\varepsilon = \sum \alpha_i \phi(i)$ with $\sum \alpha_i = 1$ and $\alpha_i \geq 0$. Hence, $$ \sum_{i\leq k} \alpha_i r(i) = f_\varepsilon(k) $$ But then, \begin{align*} \| f_\varepsilon \|_\infty = \sum_{i \in \mathbb N} \alpha_i r(i) \geq \sum_{i\leq k} \alpha_i r(i) + \sum_{i \geq k} \alpha_i &= 1 + \sum_{i \leq k}\alpha_i(r(i)-1) \geq 1 + \sum_{i \leq k} \alpha_i \frac{r(i)}{k+1} \\ &= 1 + \frac{f_\varepsilon(k)}{k+1} \\ &\geq 1 +2 \varepsilon \end{align*} But this is a contradiction to $\|f_0\|_\infty \leq 1$ and $\|f_\varepsilon - f_0\| \leq \varepsilon$.

In summary, we have found a convex closed subset of $\ell_\infty(\mathbb N)$ and a sequence of nested balls with empty intersection, showing (5).

Regarding (6), the 1-Wasserstein-space of (4) seems to not work (at least when inheriting the balls), as the balls are not nested anymore. But maybe more sophisticated examples are possible?

Answer 3

Let $$ X\,\ :=\,\ (1;2)\ =\ \{x\in\mathbb R: 1< x< 2\} $$ The distance in $\ X\ $ is defined by:

$$ \forall_{x\ y\in X}\quad d(x\ y)\ :=\ \max(x\ y) $$

This ultra-metric space is discrete hence complete.

Also:

$$ \forall_{x\in X}\quad \{d(x\ y) : y\in X\}\ =\ [x;2)\ $$

Finally, the following descending sequence of nonempty closed balls $\ B(x\ r):=\{y\in X: d(x\ y)\le r\}\ $ has an empty intersection:

$$ \bigcap_{n=1}^\infty\ B\left(\frac{n+1}n\ \ \frac1n\right)\,\ =\,\ \emptyset $$ Thus, all four assumptions are satisfied.