Tag Archives: metric-spaces

Counterexamples around balls in metric spaces

Let’s play with balls in a metric space \((M,d)\). We denote by

  • \(B_r(p) = \{x \in M : d(x,p) < r\}\) the open ball.
  • \(B_r[p] = \{x \in M : d(x,p) \le r\}\) the closed ball.

A ball of radius \(r\) included in a ball of radius \(r^\prime < r\)

We take for \(M\) the space \(\{0\} \cup [2, \infty)\) equipped with the standard metric distance \(d(x,y)=\vert x – y \vert\).

We have \(B_4(0) = \{0\} \cup [2, 4)\) while \(B_3(2) = \{0\} \cup [2, 5)\). Despite having a strictly smaller radius, the ball \(B_3(2)\) strictly contains the ball \(B_4(0)\).

The phenomenon cannot happen in a normed vector space \((M, \Vert \cdot \Vert)\). For the proof, take two open balls \(B_r(p),B_{r^\prime}(p^\prime) \subset M\), \(0 < r^\prime < r\) and suppose that \(p \in B_{r^\prime}(p^\prime)\). If \(p=p^\prime\) and \(q \in B_{r^\prime}(p^\prime) \setminus \{p^\prime\}\) then \(p + \frac{\frac{r+r^\prime}{2} }{\Vert p q \Vert} p q \in B_r(p) \setminus B_{r^\prime}(p^\prime) \). And if \(p \neq p^\prime\), \(p \in B_{r^\prime}(p^\prime)\) then \(p^\prime + \frac{\frac{r+r^\prime}{2} }{\Vert p^\prime p \Vert} p^\prime p \in B_r(p) \setminus B_{r^\prime}(p^\prime) \).

An open ball \(B_r(p)\) whose closure is not equal to the closed ball \(B_r[p]\)

Here we take for \(M\) a subspace of \(\mathbb R^2\) which is the union of the origin \(\{0\}\) with the unit circle \(S^1\). For the distance, we use the Euclidean norm.
The open unit ball centered at the origin \(B_1(0)\) is reduced to the origin: \(B_1(0) = \{0\}\). Its closure \(\overline{B_1(0)}\) is itself. However the closed ball \(B_1[0]\) is the all space \(\{0\} \cup S^1\).

Again one can prove that for a normed vector space this cannot happen. The closure of an open ball is the closed ball for a normed vector space.

Counterexamples to Banach fixed-point theorem

Let \((X,d)\) be a metric space. Then a map \(T : X \to X\) is called a contraction map if it exists \(0 \le k < 1\) such that \[d(T(x),T(y)) \le k d(x,y)\] for all \(x,y \in X\). According to Banach fixed-point theorem, if \((X,d)\) is a complete metric space and \(T\) a contraction map, then \(T\) admits a fixed-point \(x^* \in X\), i.e. \(T(x^*)=x^*\).

We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.

First, let’s consider \[\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
& x & \longmapsto & x+1 \end{array}\] For all \(x,y \in \mathbb R\) we have \(\vert f(x)-f(y) \vert = \vert x- y \vert\). \(f\) is not a contraction, but an isometry. Obviously, \(f\) has no fixed-point.

We now prove that a map satisfying \[d(g(x),g(y)) < d(x,y)\] might also not have a fixed-point. A counterexample is the following map \[\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}\] Since \[g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),\] by the mean value theorem \[\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).\] However \(g\) has no fixed-point. Finally, let's have a look to a space \((X,d)\) which is not complete. We take \(a,b \in \mathbb R\) with \(0 < a < 1\) and for \((X,d)\) the space \(X = \mathbb R \setminus \{\frac{b}{1-a}\}\) equipped with absolute value distance. \(X\) is not complete. Consider the map \[\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}\] \(h\) is well defined as for \(x \neq \frac{b}{1-a}\), \(h(x) \neq \frac{b}{1-a}\). \(h\) is a contraction map as for \(x,y \in \mathbb R\) \[\vert h(x)-h(y) \vert = a \vert x - y \vert \] However, \(h\) doesn't have a fixed-point in \(X\) as \(\frac{b}{1-a}\) is the only real for which \(h(x)=x\).

A separable space that is not second-countable

In topology, a second-countable space (also called a completely separable space) is a topological space having a countable base.

It is well known that a second-countable space is separable. For the proof consider a second-countable space \(X\) with countable basis \(\mathcal{B}=\{B_n; n \in \mathbb{N}\}\). We can assume without loss of generality that all the \(B_n\) are nonempty, as the empty ones can be discarded. Now, for each \(B_n\), pick any element \(b_n\). Let \(D=\{b_n;n \in \mathbb{N}\}\). \(D\) is countable. We claim that \(D\) is dense in \(X\). To see this let \(U\) be any nonempty open subset of \(X\). \(U\) contains some \(B_p\), hence \(b_p \in U\). So \(D\) intersects \(U\) proving that \(D\) is dense.

What about the converse? Is a separable space second-countable? The answer is negative and I present below a counterexample. Continue reading A separable space that is not second-countable