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Let $x,y\in \mathbb R$.

A common definition of Euclidean line segment $xy$ on $\mathbb R$ - real line - going from $x$ to $y$ is the set ${\{s:s=tx+(1-t)y:0 \leq t \leq 1}\}$.

Obviously, since we aren't considering directed line segments, but rather simply "line segments", the segment $xy$ is equal to the segment $yx$.

I can't, however, deduce the above from the definition - how to prove that ${\{tx+(t-1)y:0 \leq t \leq 1}\}$ is equal to ${\{ty+(t-1)x:0 \leq t \leq 1}\}$?

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    Use the change of variable $t'=1-t$ – 2017-02-08

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$ 0 \le t \le 1 \iff 0 \le 1-t \le 1$. And $t = 1-(1-t)$

So $xy = {\{s:s=tx+(1-t)y:0 \leq t \leq 1}\}$

$ = {\{s:s=(1-(1-t))x+(1-t)y:0 \leq 1-t \leq 1}\}$

$={\{s:s=uy+(1-u)x:0 \leq u \leq 1: u = 1-t}\}$

$={\{s:s=ty+(1-t)x:0 \leq t \leq 1}\}=yx$ .