An open set containing a closed ball should contain an open ball between them

Question

In this question I am probably missing some easy counterexamples:

Let $(X,d)$ be a metric space, fix $x_0\in X$, and suppose that an open set $G$ contains the closed ball $F:=\{x \in X: d(x,x_0)\le 1\}$. Does there exist an open ball $B$ centered in $x_0$ such that $F\subseteq B\subseteq G$?

Idea for a possible counterexample: Pick an infinite dimensional Banach space $X$, set $x_0=0$, and fix infinitely many points $z_i$ on the unit sphere (the boundary of $F$). Then construct $G$ as the union between $F$ and all the open balls with centers $z_i$ and radii $r_i$ with $\inf_i r_i=0$ (the union of the latter open balls should contain the unit sphere).

Answer 1

This is just a variation on the proposed counter-example (with details). Let $X$ be an infinite-dimensional real Banach vector space, e.g. the Hilbert space $\ell_2$. The unit sphere $S=\{x: |x|=1\}\subset X$ is noncompact. Hence, it admits a bounded continuous function $f: S\to {\mathbb R}$ whose infimum is $1$ but the infimum is not attained (this is a consequence of Tietze's extension theorem if you like). Consider the spherical coordinates $y=(x,r)$ on $X_0:=X\setminus \{0\}$, $y=rx$, $x\in S, r\in (0,\infty)$. On $X_0$ take $$ G_0=\{y: r<f(x)\}. $$ Lastly, set $G:= G_0\cup \{0\}$, an open subset of $X$. Then $G$ contains the unit ball $F=\{y: |y|\le 1\}$. But there is no open ball between $F$ and $G$.

Answer 2

This is not an answer to the question, rather a guidance for any future answers.

No counterexample exists for the Euclidean spaces $\Bbb R^n$ or $\Bbb C^n$. Let $B(-,-)$ denote open balls.

Say $(X,\rho)$ is a Euclidean space, take $x_0\in X$ and $F=\{x\in X:\rho(x,x_0)\le1\}$. Say $G\supseteq F$ is open. Notice $F$ is compact.

$K:=X\setminus G$ is closed. $\rho(\cdot,K):F\to(0,\infty)$ is continuous and assumes a minimum $\delta>0$. We may omit $0$ from the image since $\rho(x,K)=0$ implies $x\in K$. If $y\in X\setminus F$ has $\rho(y,x_0)<1+\delta$ I note that, by moving along a line $y\to x_0$, we may identify a closest point $\alpha\in F$ to $y$ with $\rho(\alpha,x_0)=1$. Because the triangle inequality is an equality for points on the same line, we deduce that $\rho(\alpha,y)<\delta$ hence $y\in G$. Thus: $$F\subseteq B(x_0,1+\delta)\subseteq G$$

Answer 3

Assume that the following two properties are satisfied:

1. The sphere $S_X = S_X(x_0) = \{ x \in X \mid d(x,x_0) = 1 \}$ is not compact.

2. For each $x \in S_X$ and each $n \in \mathbb N$ there exists $x' \in X \setminus F$ such that $d(x',x) < 1/n$.

Then there exists an open neigborhood $G$ of $F$ in $X$ which does not admit an open ball $B = B(x_0,r) = \{x \in X \mid d(x,x_0) < r \}$ such that $F \subset B \subset G$.

Proof. Since $S_X$ is not compact, there exists a sequence $(x_n)$ in $S_X$ which has no convergent subsequence. Pick $x'_n \in X \setminus F$ such that $d(x'_n,x_n) < 1/n$. Let $A = \{x'_n \mid n \in \mathbb N\}$. Clearly $$F \subset G = X \setminus A.$$

• We claim that $A$ is closed in $X$ (i.e. that $G$ is an open neigborhood of $F$ in $X$).
  Assume that $A$ is not closed in $X$. Then there exists $\xi \in X \setminus A$ and a sequence $(\xi_k)$ in $A$ such that $\xi_k \to \xi$. Write $\xi_k = x'_{n_k}$. The sequence $(n_k)$ cannot be bounded (otherwise all $\xi_k$ would be contained in a finite subset $A' \subset A$ which would imply $\xi \in A'$). Thus we may assume w.l.o.g. that $(n_k)$ is strictly increasing. But then $x_{n_k} \to \xi$ which is a contradiction: In fact, for each $\epsilon > 0$ we get $d(x_{n_k},\xi) \le d(x_{n_k},x'_{n_k}) + d(x'_{n_k},\xi) < 1/n_k + d(\xi_k,\xi) < \epsilon$ for $k \ge k_0$.

Now assume that $F \subset B(x_0,r) \subset G$ for some $r$.

• We claim that we must have $r > 1$.
  Assume that $F \subset B(x_0,r)$ for some $r \le 1$. Then certainly $F \subset B(x_0,1)$. Since $S_X \subset F$ and $S \cap B(x_0,1) = \emptyset$, we conclude $S_X = \emptyset$ which is compact, a contradiction.

• $B(x_0,r) \subset G$ means that $B(x_0,r) \cap A = \emptyset$. Pick $n$ such that $1 + 1/n < r$. We have $x'_n \in A$ and $d(x'_n,x_n) < 1/n$, hence $d(x'_n,x_0) \le d(x'_n,x_n) + d(x_n,x_0) < 1/n + 1 < r$, thus $x'_n \in B(x_0,r)$ which is a contradiction.

Let us now give some examples.

Let $(V, \lVert - \rVert)$ be a normed linear space. It is a metric space via $d(v,w) = \lVert v - w \rVert$. Each subset $X \subset V$ inherits a metric from $V$ which will again be denoted by $d$.

Let $S = S_V(0)$ denote the sphere in $V$ with center $0$ and radius $1$. Take any non-compact subset $S_0 \subset S$ and define $X = \{ t \cdot x \mid t \in [0,\infty), x \in S_0 \}$. Take $x_0 = 0$. Then 1. is satisfied because $S_X = S_0$ and 2. is trivially satisfied by taking $x' = (1 +1/2n)x$.

This covers Moishe Kohan's answer: In an infinite-dimensional normed linear space $V$ the sphere $S$ is never compact, thus we may take $X = V$ and $x_0 = 0$.

In $V = \mathbb R^n$ we may take $X = V \setminus \{x\}$ for any $x \in S$. In that case we can argue more directly than with a sequence $(x'_n)$ as in the above proof. Simply take $G = B(-x,2)$.

Update:

There are also counterexamples in case $S_X$ (or even $F$) is compact.

Write points $x \in \mathbb R^n$ as $x = (x_1,\ldots, x_n)$ and define subsets $A, B, H, X \subset \mathbb R^n$ by $$A = \{ x \in \mathbb R^n \mid \lVert x \rVert \le 1 , x_n \ge 0\} = \text{ closed upper unit halfball in } \mathbb R^n,$$ $$B = \{ x \in \mathbb R^n \mid \lVert x \rVert > 1 \} = \text{ exterior of the closed unit ball} ,$$ $$H = \{ x \in \mathbb R^n \mid x_n < -1 \} = \text{ open halfspace with } x_n < - 1 ,$$ $$X = A \cup B .$$ For $x_0 = 0$ we get $S_X = \{ x \in \mathbb R^n \mid \lVert x \rVert = 1 , x_n \ge 0\}$ and $F = A$ which are both compact sets. We have $H \subset B \subset X$. Let $N =(0,\ldots,0,1)$ be the north pole of $A$.

The open ball $B(N,2)$ in $X$ clearly contains $F = A$. We claim that there does not exist an open ball $B(0,r)$ such that $F \subset B(0,r) \subset B(N,2)$.

Let $r > 0$ such that $F \subset B(0,r)$. This is possible only when $r > 1$. The ball $B(N,2)$ does not intersect $H$, but the ball $B(0,r)$ does. Thus it is impossible that $B(0,r) \subset B(N,2)$.

Answer 4

A counterexample that just works in a subset of $\mathbb R^2$.

Use $X= \{(x,y) \in \mathbb R^2: y >0\}$, the distance function $d$ being the same as in $\mathbb R^2$. $G = \{(x,y) \in X: y > \frac12x^2\}$ is an open subset of $(X,d)$. If we choose $x_0=(0,1) \in X$, then $F$ is just the normal closed unit ball around $x_0$ in $\mathbb R^2$, except for the point $(0,0)$, which is not in $X$.

We have $F \subseteq G$. The easiest way to see this, arguing geometrically in $\mathbb R^2$, is that $x_0$ is in $G$, so we need to find out what the smallest distance from $x_0$ to any point on the boundary of $G$ is, which is the function $y=\frac12x^2$.

We have

$$d^2\left((0,1), (x, \frac12x^2)\right) = x^2 + (1-\frac12x^2)^2 = \frac14x^4+1.$$

That (squared) distance is bigger than $1$ for any $x \in \mathbb R$ except for $x=0$, but in this cases the corresponding point $(0,0)$ is not in $X$. That means that we indeed have $F \subseteq G$.

What matters here is that (looking at $\mathbb R^2$ again) both the lower boundary of $F$ (the function $f(x)=1-\sqrt{1-x^2})$ and of $G$ ($g(x)=\frac12x^2)$ are continuous, go through $(0,0)$, are positive except at that point and that $g(x) < f(x)$ except at $x=0$.

Plotted this looks like this (I had to use $\frac1{20}$ as factor to get a visible gap between the 2 functions, otherwise they are just very close together near $0$):

Why is this configuration a counterexample to the claim presented in the problem statement?

Since $F$ contains many points with distance $1$ to $x_0$ (e.g. $(0,2)$), any open ball $B(x_0,r)$ with $F \subseteq B$ needs to have $r > 1$.

But for any $r>1$ there is an $x_r>0$ with $1+x_r^2=r^2$. Looking at the line segement from $(0,1)$ to $(x_r,0)$, we see that $B$ contains it completely except for its endpoint $(x_r,0)$. But the above picture makes it clear that that line segment intersects the red curve (=lower boundary of $G$) at some point. Since the red curve never meets the $x$-axis except at $x=0$, that means this intersection point is in $B$, but not in $G$ (which contains only points above the red curve).

This concludes the proof, any open ball $B$ wanting to enclose $F$ will necessarily contain points not in $G$.