Valid Transitive Closure?

Question

Is the relation $D_r\cup\{(x,z)\in\mathbf{R}\times\mathbf{R}|\exists y\in\mathbf{R}(\|x-y\|+\|y-z\|<r)\}$ the transitive closure to the relation $D_r = \{(x,y)\in\mathbf{R}\times\mathbf{R}|\land\|x-y\|<r\}$ where $r\in\mathbf{R}$ and $r>0$ ?

Answer 1

No, because the relation $\mathbb R$ is not transitive:

$$\mathbb R = \{(x,z)\in\mathbf{R}\times\mathbf{R} | \exists y\in\mathbf{R} (|x-y|+|y-z|<r)\}$$

An intuitive example to show that is below. Let's assume that $r = 2.5$, $a = 1$, $b = 3$ and $c = 5$.

The $(a, b) \in \mathbb R$, cause you can find a (middle) point $m_1 = 2$ between points $a$ and $b$, such that:

$$|a-m_1| + |m_1-b| < r$$

The $(b, c) \in \mathbb R$ as well, cause you can find a (middle) point $m_2 = 4$ between points $b$ and $c$, such that:

$$|b-m_2| + |m_2-c| < r$$

However, $(a, c) \notin \mathbb R$, because the distance between points $a$ and $c$ is $4$, so it'll be impossible to find a (middle) point $m_3$, such that:

$$|a-m_3| + |m_3-c| < r$$

QED