# 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