Definition of Convex Set The definition given in my textbook for a convex set is; "Let $E$ be a subset of $\Bbb R^n$. We say $E$ is convex if given $x,y \in E$, $(1-t)x +ty \in E$ for all real numbers $t$ with $0\le t\le 1$."
My doubt are:  


*

*Technically is the definition also valid if we take $tx+(1-t)y \in E$ instead of $(1-t)x + ty \in E$? To me it appears so. Is it just convention or is there a catch that I am not able to see?

*Since $E$ is subset of $\Bbb R^n$ why are we considering only $x$ and $y$ (because then shouldn't this definition be for $\Bbb R^2$)?
 A: Yes, if you substitute $t$ with $1-t$, you arrive at $(1-t)x+ty$ instead of $tx+(1-t)y$, and $0\le 1-t\le 1$ is equivalent to $0\le t\le 1$.
A more symmetric way of definingt convexity might be to demand $tx+uy\in E$ for all $t,u\ge0$ with $t+u=1$.
Here, $x$ and $y$ are not the two coordinates of a single point $(x,y)\in \Bbb R^2$. Instead, they are elements of $E$ (and hence of $\Bbb R^n$.
If you wish, view $x$ as $(x_1,\ldots, x_n)$ and $y$ as $(y_1,\ldots, y_n)$ and write $tx+(1-t)y$ clumsily as $\bigl(tx_1+(1-t)y_1,\ldots, tx_n+(1-z)y_n\bigr)$.
A: A convex set is a set with the property, said in words:
If you take two points in this set, then the straight line connecting them lies in the set.
This is the answer to one of your questions, why you only take two points $x$ and $y$. These are the two arbitrary points you take as the ends of your straight line.
The other point is that is is certainly equivalent if you say
$$ tx + (1-t)y \in E \ \forall t \in [0,1] $$ or
$$ (1-t)x + ty \in E \ \forall t \in [0,1]. $$ 
Why is this equivalent? Well, because if you have the first property and take some $t \in [0,1]$, then just define $s = (1-t)$ and for $s$ we then have $ (1-s)x + sy \in E$. In this way, you can transform one description into the other. 
