# A connected not locally connected space

In this article, I will describe a subset of the plane that is a connected space while not locally connected nor path connected.

Let’s consider the plane $$\mathbb{R}^2$$ and the two subspaces:
$A = \bigcup_{n \ge 1} [(0,0),(1,\frac{1}{n})] \text{ and } B = A \cup (\frac{1}{2},1]$ Where a segment noted $$|a,b|$$ stands for the plane segment $$|(a,0),(b,0)|$$. Continue reading A connected not locally connected space

# Continuous maps that are not closed or not open

We recall some definitions on open and closed maps. In topology an open map is a function between two topological spaces which maps open sets to open sets. Likewise, a closed map is a function which maps closed sets to closed sets.

For a continuous function $$f: X \mapsto Y$$, the preimage $$f^{-1}(V)$$ of every open set $$V \subseteq Y$$ is an open set which is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in $$Y$$ are closed in $$X$$. However, a continuous function might not be an open map or a closed map as we prove in following counterexamples. Continue reading Continuous maps that are not closed or not open

# Two disjoint closed sets with distance equal to zero

We take a metric space $$(E,d)$$ and consider two closed subsets $$A,B$$ having a distance $$d(A,B)$$ equal to zero. We raise the following question: can $$A$$ and $$B$$ be disjoint – $$A \cap B=\emptyset$$? Continue reading Two disjoint closed sets with distance equal to zero

# A curve filling a square – Lebesgue example

## Introduction

We aim at defining a continuous function $$\varphi : [0,1] \rightarrow [0,1]^2$$. At first sight this looks quite strange.

Indeed, $$\varphi$$ cannot be a bijection. If $$\varphi$$ would be bijective, it would also be an homeomorphism as a continuous bijective function from a compact space to a Haussdorff space is an homeomorphism. But an homeomorphism preserves connectedness and $$[0,1] \setminus \{1/2\}$$ is not connected while $$[0,1]^2 \setminus \{\varphi(1/2)\}$$ is.

Nor can $$\varphi$$ be piecewise continuously differentiable as the Lebesgue measure of $$\varphi([0,1])$$ would be equal to $$0$$.

$$\varphi$$ is defined in two steps using the Cantor space $$K$$. Continue reading A curve filling a square – Lebesgue example

# Cantor set: a null set having the cardinality of the continuum

## Definition of the Cantor set

The Cantor ternary set (named Cantor set below) $$K$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$K_0=I$$
• If $$K_n$$ is a finite disjoint union of segments $$K_n=\cup_k \left[a_k,b_k\right]$$, $K_{n+1}=\bigcup_k \left(\left[a_k,a_k+\frac{b_k-a_k}{3}\right] \cup \left[a_k+2\frac{b_k-a_k}{3},b_k\right]\right)$

And finally $$K=\displaystyle \bigcap_{n \in \mathbb{N}} K_n$$. The Cantor set is created by repeatedly deleting the open middle third of a set of line segments starting with the segment $$I$$.

The Cantor set is a closed set as it is an intersection of closed sets. Continue reading Cantor set: a null set having the cardinality of the continuum

# An unbounded convex not containing a ray

We consider a normed vector space $$E$$ over the field of the reals $$\mathbb{R}$$ and a convex subset $$C \subset E$$.

We suppose that $$0 \in C$$ and that $$C$$ is unbounded, i.e. there exists points in $$C$$ at distance as big as we wish from $$0$$.

The following question arises: “does $$C$$ contains a ray?”. It turns out that the answer depends on the dimension of the space $$E$$. If $$E$$ is of finite dimension, then $$C$$ always contains a ray, while if $$E$$ is of infinite dimension $$C$$ may not contain a ray. Continue reading An unbounded convex not containing a ray

# A compact whose convex hull is not compact

We consider a topological vector space $$E$$ over the field of the reals $$\mathbb{R}$$. The convex hull of a subset $$X \subset E$$ is the smallest convex set that contains $$X$$.

The convex hull may also be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X.

The convex hull of $$X$$ is written as $$\mbox{Conv}(X)$$. Continue reading A compact whose convex hull is not compact

# A compact convex set whose extreme points set is not close

Let’s remind that an extreme point $$c$$ of a convex set $$C$$ in a real vector space $$E$$ is a point in $$C$$ which does not lie in any open line segment joining two points of $$C$$.

## The specific case of dimension $$2$$

Proposition: when $$C$$ is closed and its dimension is equal to $$2$$, the set $$\hat{C}$$ of its extreme points is closed.
Continue reading A compact convex set whose extreme points set is not close

# An empty intersection of nested closed convex subsets in a Banach space

We consider a decreasing sequence $$(C_n)_{n \in \mathbb{N}}$$ of non empty closed convex subsets of a Banach space $$E$$.

If the convex subsets are closed balls, their intersection is not empty. To see this let $$x_n$$ be the center and $$r_n > 0$$ the radius of the ball $$C_n$$. For $$0 \leq n < m$$ we have $$\Vert x_m-x_n\Vert \leq r_n – r_m$$ which proves that $$(x_n)_{n \in \mathbb{N}}$$ is a Cauchy sequence. As the space $$E$$ is Banach, $$(x_n)_{n \in \mathbb{N}}$$ converges to a limit $$x$$ and $$x \in \bigcap_{n=0}^{+\infty} C_n$$. Continue reading An empty intersection of nested closed convex subsets in a Banach space