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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?

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    @LeeMosher I don't have any idea where to start. Considering that I can take different points and create the same straight line, I suppose it has something to do with the fact that the value of $t$ is bounded by $0$ and $1$, but I don't know how to use that information. – 2017-02-12

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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?

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    I suppose you are talking about the fact that one can get the values of the endpoints by setting $t=0$ or $t=1$. Is there a proof that $\forall t: ta+(1-t)b=tc+(1-t)d$? If not, how to expand on this result? – 2017-02-12
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    No, there is no such proof. For example, setting $t=1$ you would get $a=c$ which is not necessarily true. And that's a good thing, because you are trying to prove that $\overline{ab}=\overline{cd} \iff \{a,b\}=\{c,d\}$. – 2017-02-12
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    Is the fact that $a,b \in {\{tc+(1-t)d}\}$ of any use here? – 2017-02-12
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    Not so sure that's a fact. For instance, if $a=0$, $b=1$, $c=-1$, $d=2$ then it's false. – 2017-02-13
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    Then we know that these numbers can't be the values of $a,b,c,d$. We know that $a,b \in {\{ta+(1-t)b}\}$, the result above comes from ${\{ta+(1-t)b}\}={\{tc+(1-t)d}\}$. Do you have some more concrete clues? – 2017-02-13
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    At this point, the best I can do to help you is to repeat my earlier suggestion: To answer your original question, prove that for any $a \ne b$, $c \ne d \in \mathbb{R}^n$, the segments $\overline{ab}=\{(1-t)a + tb \,|\, 0 \le t \le 1\}$ and $\overline{cd}=\{(1-t)c+td \,|\, 0 \le t \le 1\}$ are equal if and only if the sets $\{a,b\}$, $\{c,d\}$ are equal. – 2017-02-13
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    Well, you've already told me that before. It's trivial to prove that if the latter sets are equal then the former are, I want to know this statement's converse. My problem is that I don't even know where to start with proving it. How do I prove that if the segments are equal, ${\{a,b}\}$ and ${\{c,d}\}$ are equal too? – 2017-02-13