For a non-empty compact set C ⊂ R (or $R^2$) prove that there is the point x ∈ C such that y <lex x for any y ∈ C\ {x}.
From the analysis, we know that supremum of any subset of R exits. And since C is a compact set (i.e it is bounded and closed) therefore its supremum belongs to C. Let x = sup C. Then there is a sequence of number i.e. $y_1,y_2,..y_n$ ∈ C. So the limit $y_i$ = x, and due to the closeness x ∈ C. This is what I have tried in case of R but I am not sure if it's correct or do I have to include something else and am unable to do so for $R^2$. Any kind of feedback is appreciated?
Note: '<lex' denotes 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]
Thanks in advance.
HINT: Let $\pi_1:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$ and $\pi_2:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto y$ be the projection maps to the coordinate axes. Show that $\pi_1[C]$ has a maximum element, say $a$. Then show that $\{y\in\Bbb R:\langle a,y\rangle\in C\}$ has a maximum element, say $b$. Finally, show that $\langle a,b\rangle$ is the maximum element of $C$ in the lexicographic order on $\Bbb R^2$.