A non complete normed vector space

Consider a real normed vector space $$V$$. $$V$$ is called complete if every Cauchy sequence in $$V$$ converges in $$V$$. A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like $$(\ell_p, \Vert \cdot \Vert_p)$$ the space of real sequences endowed with the norm $$\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}$$ for $$p \ge 1$$, the space $$(C(X), \Vert \cdot \Vert)$$ of real continuous functions on a compact Hausdorff space $$X$$ endowed with the norm $$\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert$$ or the Lebesgue space $$(L^1(\mathbb R), \Vert \cdot \Vert_1)$$ of Lebesgue real integrable functions endowed with the norm $$\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx$$.

Let’s give an example of a non complete normed vector space. Let $$(P, \Vert \cdot \Vert_\infty)$$ be the normed vector space of real polynomials endowed with the norm $$\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert$$. Consider the sequence of polynomials $$(p_n)$$ defined by
$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.$ For $$m < n$$ and $$x \in [0,1]$$, we have $\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}$ which proves that $$(p_n)$$ is a Cauchy sequence. Also for $$x \in [0,1]$$ $\lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.$ As uniform converge implies pointwise convergence, if $$(p_n)$$ was convergent in $$P$$, it would be towards $$p$$. But $$p$$ is not a polynomial function as none of its $$n$$th-derivative always vanishes. Hence $$(p_n)$$ is a Cauchy sequence that doesn't converge in $$(P, \Vert \cdot \Vert_\infty)$$, proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

Intersection and union of interiors

Consider a topological space $$E$$. For subsets $$A, B \subseteq E$$ we have the equality $A^\circ \cap B^\circ = (A \cap B)^\circ$ and the inclusion $A^\circ \cup B^\circ \subseteq (A \cup B)^\circ$ where $$A^\circ$$ and $$B^\circ$$ denote the interiors of $$A$$ and $$B$$.

Let’s prove that $$A^\circ \cap B^\circ = (A \cap B)^\circ$$.

We have $$A^\circ \subseteq A$$ and $$B^\circ \subseteq B$$ and therefore $$A^\circ \cap B^\circ \subseteq A \cap B$$. As $$A^\circ \cap B^\circ$$ is open we then have $$A^\circ \cap B^\circ \subseteq (A \cap B)^\circ$$ because $$A^\circ \cap B^\circ$$ is open and $$(A \cap B)^\circ$$ is the largest open subset of $$A \cap B$$.

Conversely, $$A \cap B \subseteq A$$ implies $$(A \cap B)^\circ \subseteq A^\circ$$ and similarly $$(A \cap B)^\circ \subseteq B^\circ$$. Therefore we have $$(A \cap B)^\circ \subseteq A^\circ \cap B^\circ$$ which concludes the proof of the equality $$A^\circ \cap B^\circ = (A \cap B)^\circ$$.

One can also prove the inclusion $$A^\circ \cup B^\circ \subseteq (A \cup B)^\circ$$. However, the equality $$A^\circ \cup B^\circ = (A \cup B)^\circ$$ doesn’t always hold. Let’s provide a couple of counterexamples.

For the first one, let’s take for $$E$$ the plane $$\mathbb R^2$$ endowed with usual topology. For $$A$$, we take the unit close disk and for $$B$$ the plane minus the open unit disk. $$A^\circ$$ is the unit open disk and $$B^\circ$$ the plane minus the unit closed disk. Therefore $$A^\circ \cup B^\circ = \mathbb R^2 \setminus C$$ is equal to the plane minus the unit circle $$C$$. While we have $A \cup B = (A \cup B)^\circ = \mathbb R^2.$

For our second counterexample, we take $$E=\mathbb R$$ endowed with usual topology and $$A = \mathbb R \setminus \mathbb Q$$, $$B = \mathbb Q$$. Here we have $$A^\circ = B^\circ = \emptyset$$ thus $$A^\circ \cup B^\circ = \emptyset$$ while $$A \cup B = (A \cup B)^\circ = \mathbb R$$.

The union of the interiors of two subsets is not always equal to the interior of the union.

The line with two origins

Let’s introduce and describe some properties of the line with two origins.

Let $$X$$ be the union of the set $$\mathbb R \setminus \{0\}$$ and the two-point set $$\{p,q\}$$. The line with two origins is the set $$X$$ topologized by taking as base the collection $$\mathcal B$$ of all open intervals in $$\mathbb R$$ that do not contain $$0$$, along with all sets of the form $$(-a,0) \cup \{p\} \cup (0,a)$$ and all sets of the form $$(-a,0) \cup \{q\} \cup (0,a)$$, for $$a > 0$$.

$$\mathcal B$$ is a base for a topology $$\mathcal T$$ of $$X$$

Indeed, one can verify that the elements of $$\mathcal B$$ cover $$X$$ as $X = \left( \bigcup_{a > 0} (-a,0) \cup \{p\} \cup (0,a) \right) \cup \left( \bigcup_{a > 0} (-a,0) \cup \{q\} \cup (0,a) \right)$ and that the intersection of two elements of $$\mathcal B$$ is the union of elements of $$\mathcal B$$ (verification left to the reader).

Each of the spaces $$X \setminus \{p\}$$ and $$X \setminus \{q\}$$ is homeomorphic to $$\mathbb R$$

Let’s prove it for $$X \setminus \{p\}$$. The map $\begin{array}{l|rcll} f : & X \setminus \{p\} & \longrightarrow & \mathbb R \\ & x & \longmapsto & x & \text{for } x \neq q\\ & q & \longmapsto & 0 \end{array}$ is a bijection. $$f$$ is continuous as the inverse image of an open interval $$I$$ of $$\mathbb R$$ is an open subset of $$X$$. For example taking $$I=(-b,c)$$ with $$0 < b < c$$, we have \begin{align*} f^{-1}[I] &= (-b,0) \cup \{q\} \cup (0,c)\\ &= \left( (-b,0) \cup \{q\} \cup (0,b) \right) \cup (b/2,c) \end{align*} One can also prove that $$f^{-1}$$ is continuous. Continue reading The line with two origins

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.

The Smith Volterra Cantor Set

In Cantor set article, I presented the Cantor set which is a null set having the cardinality of the continuum. I present here a modification of the Cantor set named the Smith-Volterra-Cantor set.

Construction of the Smith-Volterra-Cantor set

The Smith-Volterra-Cantor set (also named SVC set below) $$S$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$S_0=I$$
• $$S_1=[0,\frac{3}{8}] \cup [\frac{5}{8},1]$$
• If $$S_n$$ is a finite disjoint union of segments $$s_n=\cup_k \left[a_k,b_k\right]$$, $S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)$

Counterexamples around connected spaces

A connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. We look here at unions and intersections of connected spaces.

Union of connected spaces

The union of two connected spaces $$A$$ and $$B$$ might not be connected “as shown” by two disconnected open disks on the plane.

However if the intersection $$A \cap B$$ is not empty then $$A \cup B$$ is connected.

Intersection of connected spaces

The intersection of two connected spaces $$A$$ and $$B$$ might also not be connected. An example is provided in the plane $$\mathbb R^2$$ by taking for $$A$$ the circle centered at the origin with radius equal to $$1$$ and for $$B$$ the segment $$\{(x,0) \ : \ x \in [-1,1]\}$$. The intersection $$A \cap B = \{(-1,0),(1,0)\}$$ is the union of two points which is not connected.

Playing with interior and closure

Let’s play with the closure and the interior of sets.

To start the play, we consider a topological space $$E$$ and denote for any subspace $$A \subset E$$: $$\overline{A}$$ the closure of $$A$$ and $$\overset{\circ}{A}$$ the interior of $$A$$.

Warm up with the closure operator

For $$A,B$$ subsets of $$E$$, the following results hold: $$\overline{\overline{A}}=\overline{A}$$, $$A \subset B \Rightarrow \overline{A} \subset \overline{B}$$, $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$ and $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$.

Let’s prove it.
$$\overline{A}$$ being closed, it is equal to its closure and $$\overline{\overline{A}}=\overline{A}$$.

Suppose that $$A \subset B$$. As $$B \subset \overline{B}$$, we have $$A \subset \overline{B}$$. Also, $$\overline{B}$$ is closed so it contains $$\overline{A}$$, which proves $$\overline{A} \subset \overline{B}$$.

Let’s consider $$A,B \in E$$ two subsets. As $$A \subset A \cup B$$, we have $$\overline{A} \subset \overline{A \cup B}$$ and similarly $$\overline{B} \subset \overline{A \cup B}$$. Hence $$\overline{A} \cup \overline{B} \subset \overline{A \cup B}$$. Conversely, $$A \cup B \subset \overline{A} \cup \overline{B}$$ and $$\overline{A} \cup \overline{B}$$ is closed. So $$\overline{A \cup B} \subset \overline{A} \cup \overline{B}$$ and finally $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$.

Regarding the inclusion $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$, we notice that $$A \cap B \subset \overline{A} \cap \overline{B}$$ and that $$\overline{A} \cap \overline{B}$$ is closed to get the conclusion.

However, the implication $$\overline{A} \subset \overline{B} \Rightarrow A \subset B$$ doesn’t hold. For a counterexample, consider the space $$E=\mathbb R$$ equipped with the topology induced by the absolute value distance and take $$A=[0,1)$$, $$B=(0,1]$$. We have $$\overline{A}=\overline{B}=[0,1]$$.

The equality $$\overline{A} \cap \overline{B} = \overline{A \cap B}$$ doesn’t hold as well. For the proof, just consider $$A=[0,1)$$ and $$B=(1,2]$$. Continue reading Playing with interior and closure

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 $$h(x)=x$$.

Two algebraically complemented subspaces that are not topologically complemented

We give here an example of a two complemented subspaces $$A$$ and $$B$$ that are not topologically complemented.

For this, we consider a vector space of infinite dimension equipped with an inner product. We also suppose that $$E$$ is separable. Hence, $$E$$ has an orthonormal basis $$(e_n)_{n \in \mathbb N}$$.

Let $$a_n=e_{2n}$$ and $$b_n=e_{2n}+\frac{1}{2n+1} e_{2n+1}$$. We denote $$A$$ and $$B$$ the closures of the linear subspaces generated by the vectors $$(a_n)$$ and $$(b_n)$$ respectively. We consider $$F=A+B$$ and prove that $$A$$ and $$B$$ are complemented subspaces in $$F$$, but not topologically complemented. Continue reading Two algebraically complemented subspaces that are not topologically complemented