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.