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Let $\mathbb R^3$ be the set ${\{(r_1,r_2,r_3):r_1,r_2,r_3 \in \mathbb R}\}$.

Let $d:\mathbb R^3 \times \mathbb R^3 \rightarrow \mathbb R$ be a function defined by $d(x,y)=\sqrt {(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$ where $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$.

The structure $(\mathbb R^3,d)$ is then called the Euclidean space.

How are line segments defined in it?

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    Any set of the form $\{ta+(1-t)b: 0\le t \le 1\}$, where $a,b \in \Bbb R^3$. – 2017-02-08

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The segment $\overline {xy}$ is defined as the set of points $\left \{s \in \Bbb R^3:s=tx+(1-t)y, t\in \Bbb R, 1\ge t \ge0 \right \}$. This works in $\Bbb R^n$ $\forall n \in \Bbb N$.

Why is it defined like that? We are just bounding from $x$ to $y$ the only straight line for $x$ and $y$. In fact if $t=1$ then we have $x$, for $t=0$ we have $y$, otherwise we have a point on the straight line between $x$ and $y$.

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    How does this model the intuitive notion of line segment? I can't see the similarity between it and this definition. – 2017-02-08
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    I'll edit my answer, because at the moment I forgot to define it as a set of point of $\Bbb R^3$. – 2017-02-08
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    What about $x:a \le x \le b$? By definition of line passing through $a$ and $b$ as $ax+b$, why don't we make a line segment by restricting the domain of said function to the set defined above? – 2017-02-08
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    How do you define $\le$ in $\Bbb R^3$? – 2017-02-08