# Totally disconnected compact set with positive measure

Let’s build a totally disconnected compact set $$K \subset [0,1]$$ such that $$\mu(K) >0$$ where $$\mu$$ denotes the Lebesgue measure.

In order to do so, let $$r_1, r_2, \dots$$ be an enumeration of the rationals. To each rational $$r_i$$ associate the open interval $$U_i = (r_i – 2^{-i-2}, r_i + 2^{-i-2})$$. Then take $\displaystyle V = \bigcup_{i=1}^\infty U_i \text{ and } K = [0,1] \cap V^c.$ Clearly $$K$$ is bounded and closed, therefore compact. As Lebesgue measure is subadditive we have $\mu(V) \le \sum_{i=1}^\infty \mu(U_i) \le \sum_{i=1}^\infty 2^{-i-1} = 1/2.$ This implies $\mu(K) = \mu([0,1]) – \mu([0,1] \cap V) \ge 1/2.$ In a further article, we’ll build a totally disconnected compact set $$K^\prime$$ of $$[0,1]$$ with a predefined measure $$m \in [0,1)$$.

# 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

# Isometric versus affine

Throughout this article we let $$E$$ and $$F$$ denote real normed vector spaces. A map $$f : E \rightarrow F$$ is an isometry if $$\Vert f(x) – f(y) \Vert = \Vert x – y \Vert$$ for all $$x, y \in E$$, and $$f$$ is affine if $f((1-t) a + t b ) = (1-t) f(a) + t f(b)$ for all $$a,b \in E$$ and $$t \in [0,1]$$. Equivalently, $$f$$ is affine if the map $$T : E \rightarrow F$$, defined by $$T(x)=f(x)-f(0)$$ is linear.

First note that an isometry $$f$$ is always one-to-one as $$f(x) = f(y)$$ implies $0 = \Vert f(x) – f(y) \Vert = \Vert x- y \Vert$ hence $$x=y$$.

There are two important cases when every isometry is affine:

1. $$f$$ is bijective (equivalently surjective). This is Mazur-Ulam theorem, which was proven in 1932.
2. $$F$$ is a strictly convex space. Recall that a normed vector space $$(S, \Vert \cdot \Vert)$$ is strictly convex if and only if for all distinct $$x,y \in S$$, $$\Vert x \Vert = \Vert y \Vert =1$$ implies $$\Vert \frac{x+y}{2} \Vert <1$$. For example, an inner product space is strictly convex. The sequence spaces $$\ell_p$$ for $$1 < p < \infty$$ are also strictly convex.

# A non-compact closed ball

Consider a normed vector space $$(X, \Vert \cdot \Vert)$$. If $$X$$ is finite-dimensional, then a subset $$Y \subset X$$ is compact if and only if it is closed and bounded. In particular a closed ball $$B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}$$ is always compact if $$X$$ is finite-dimensional.

### The space $$A=C([0,1],\mathbb R)$$

Consider the space $$A=C([0,1],\mathbb R)$$ of the real continuous functions defined on the interval $$[0,1]$$ endowed with the sup norm:
$\Vert f \Vert = \sup\limits_{x \in [0,1]} \vert f(x) \vert$
Is the closed unit ball $$B_1[0]$$ compact? The answer is negative and we provide two proofs.

The first one is based on open covers. For $$n \ge 1$$, we denote by $$f_n$$ the piecewise linear map defined by $\begin{cases} f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\ f_n(\frac{1}{2^n})=1 \\ f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0 \end{cases}$ All the $$f_n$$ belong to $$B_1[0]$$. Moreover for $$1 \le n < m$$ we have $$\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}$$. Hence the supports of the $$f_n$$ are disjoint and $$\Vert f_n – f_m \Vert = 1$$.

Now consider the open cover $$\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}$$. For $$x \in B_1[0]\}$$ and $$u,v \in B_{\frac{1}{2}}(x)$$, $$\Vert u -v \Vert < 1$$. Therefore, each $$B_{\frac{1}{2}}(x)$$ contains at most one $$f_n$$ and a finite subcover of $$\mathcal U$$ will contain only a finite number of $$f_n$$ proving that $$A$$ is not compact.

Second proof based on convergent subsequence. As $$A$$ is a metric space, it is enough to prove that $$A$$ is not sequentially compact. Consider the sequence of functions $$g_n : x \mapsto x^n$$. The sequence is bounded as for all $$n \in \mathbb N$$, $$\Vert g_n \Vert = 1$$. If $$(g_n)$$ would have a convergent subsequence, the subsequence would converge pointwise to the function equal to $$0$$ on $$[0,1)$$ and to $$1$$ at $$1$$. As this function is not continuous, $$(g_n)$$ cannot have a subsequence converging to a map $$g \in A$$.

### Riesz’s theorem

The non-compactness of $$A=C([0,1],\mathbb R)$$ is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space $$X$$ is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of $$A=C([0,1],\mathbb R)$$ is just standard for infinite-dimensional normed vector 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.

# 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 Banach-Steinhaus theorem

In this article we look at what happens to Banach-Steinhaus theorem when the completness hypothesis is not fulfilled. One form of Banach-Steinhaus theorem is the following one.

Banach-Steinhaus Theorem
Let $$T_n : E \to F$$ be a sequence of continuous linear maps from a Banach space $$E$$ to a normed space $$F$$. If for all $$x \in E$$ the sequence $$T_n x$$ is convergent to $$Tx$$, then $$T$$ is a continuous linear map.

### A sequence of continuous linear maps converging to an unbounded linear map

Let $$c_{00}$$ be the vector space of real sequences $$x=(x_n)$$ eventually vanishing, equipped with the norm $\Vert x \Vert = \sup_{n \in \mathbb N} \vert x_n \vert$ For $$n \in \mathbb N$$, $$T_n : E \to E$$ denotes the linear map defined by $T_n x = (x_1,2 x_2, \dots, n x_n,0,0, \dots).$ $$T_n$$ is continuous as for $$\Vert x \Vert \le 1$$, we have
\begin{align*}
\Vert T_n x \Vert &= \Vert (x_1,2 x_2, \dots, n x_n,0,0, \dots) \Vert\\
& = \sup_{1 \le k \le n} \vert k x_k \vert \le n \Vert x \Vert \le n
\end{align*} Continue reading Counterexamples around Banach-Steinhaus theorem

# 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.