Let <lex denote the lexicographic ordering by their co-ordinate vertices...i.e. X <lex Y if $x_1=y_1, x_2=y_2 ... x_i< y_i.$ where X =($x_1,x_2...x_d$) and Y =($y_1,y_2...y_d$) and i ∈ [d]. Prove that for any y ∈ $R^d$ the sets {x ∈ $R^d$: x ≤lex y} and {x ∈ $R^d$: y ≤lex x} are convex.
So for proving convexity usually one assumes two points and check if tx + (1-t)y (which is basically a line segment) satisfies the given condition where t ∈ [0,1] but in lexicographic ordering, they individually should (as mentioned the coordinates) satisfy the given condition, not a line. So how can I prove them when I am not allowed to take a line segment?
Let $y=\langle y_1,\ldots,y_d\rangle$, and suppose that $x=\langle x_1,\ldots,x_d\rangle<_{\text{lex}}y$. There is a $k\in\{1,\ldots,d\}$ such that $x_i=y_i$ for $1\le i<k$, and $x_k<y_k$. Now let $t\in[0,1]$;
$$tx+(1-t)y=\big\langle tx_1+(1-t)y_1,\ldots,tx_d+(1-t)y_d\big\rangle\,,$$
and we want to show that $tx+(1-t)y\le_{\text{lex}}y$, so we consider its coordinates.
The first two bullet points are exactly what is needed to show that if $t\ne 0$, then $tx+(1-t)y<_{\text{lex}}y$, while the third shows that $tx+(1-t)y=y$ when $t=0$, so $\overline{xy}\subseteq\left\{z\in\Bbb R^d:z\le_{\text{lex}}y\right\}$.
The argument for $\left\{z\in\Bbb R^d:y\le_{\text{lex}}z\right\}$ is very similar.