For any line segment $ab$ there are unique points $a,b$

Question

Let any sets of the form ${\{s:s=ta +(1-t)b:0 \leq t \leq 1}\}$ where $a,b$ are some points in $\mathbb R^n$ be called line segments. How to prove that for each line segment, there exist unique points $a$ and $b$ - endpoints of the segment?

Answer 1

Instead of answering your question directly, let me give a hint by expanding on your guess that there is some significance to the restriction $0 \le t \le 1$.

If instead you replaced that restriction by $-\infty < t < +\infty$ then it would indeed be true that you "can take different points and create the same straight line", for example if $a=(0,0)$ and $b=(1,0)$ and $c=(-1,0)$ and $d=(2,0)$ then each of the following two lines is the $x$-axis: $$\{s : s = ta + (1-t)b \,|\, -\infty < t < +\infty\} $$ $$\{s : s = tc + (1-t)d \,|\, -\infty < t < +\infty\} $$

However, line segments are defined differently than lines themselves. In my example, the following two line segments are different: $$\{s : s = ta+(1-t)b \,|\, 0 \le t \le 1\} $$ $$\{s : s = tc + (1-t)d \,|\, 0 \le t \le 1\} $$ Now draw those two line segments and for each of them ask yourself: What are the endpoints? Can you infer a relation between the endpoints of the segment and the values of the parameter $t$? And can you generalize that inference to arbitrary line segments?