From where does $tx + (1-t)x'$ originate from? I am selfstudying an economists book, and this is popping up all of a sudden. I get that it's a line between $x$ and $x'$, but why? And is $tx' + (1-t)x$ equivalent to $tx + (1-t)x'$? I mean, if it's just the same line?
If yes, then if I take two points $x$ and $x'$ in some set $B = \{ y \ | \ y \ge b\}$ and assume without losing generality that $x' \ge x$ and then show that $tx + (1-t)x' \ge b$, is this sufficient to show that B is convex?
edit: $x$ and $x'$ are here vectors, and the $\ge$ sign is actually a preference relation.
Yes, $tx + (1-t)x'$ is a (the) straight line from the point $x$ to $x'$, for values of $t$ between $0$ and $1$.
Reversing the positions of $x$ and $x'$ makes no difference, it's still a line - it just runs in the opposite direction as you increase/decrease $t$.
And finally, yes, that would be a proof of convexity.