How to show that the distance between these sets is positive? Let $T_i=\{(1-t)x_i+ty_i;\;0\leq t\leq1\}$, where $x_i,y_i\in\mathbb{R}^n$; $i=1,2$. 
Could someone help me  to prove that if $x_2= (1+\varepsilon)x_1$ and $y_2= (1+\varepsilon)y_1$ for some $\varepsilon>0$ then $d\left (T_1,T_2\right)>0$?
Thanks.
 A: This is false: take $x_i=0$ and $y_i=0$.
A: How do you compute the distance? Is it Hausdoff distance? http://en.wikipedia.org/wiki/Hausdorff_distance
So the two sets which you have are segments $I_1, I_2$, both of them connect two points($I_i = [x_i,y_i]$). Consider the case when these segments do not coinside (i.e. the nontrivial one, when at most one pair of endpoints coinsides). Then the other does not, say, $x_1\neq x_2$. Then, if $I_1$ is not contained in $I_2$ (and we may assume without loss of generality that this is the precise case), $d(x_1,I_2) = \inf_{x\in I_2}d(x_1,x)>0$ (in fact one of the segments can be contained in another, but that doesn't change the matters...). And from here you can see that the Hausdorff distance between $I_1$ and $I_2$ is greater than 0 (cf. the link for the formula). So the statement is true whenever your segments do not coinside (they can intersect, or one can contain the other, or they may have no points in common at all)
A: I think the following holds:
If the point $0\notin T_1$ (where $0\in\mathbb{R}^n$), and $\mathbf{IF}$ the distance is calculated as below $$d(T_1,T_2)=\inf\{d(p_1,p_2):p_1\in T_1,\ p_2\in T_2\}$$ Then we get $$d(T_1,T_2)=\epsilon  \ \inf\{d(p_1,0):p_1\in T_1\}>0$$
