ukmuiik

Clash Royale CLAN TAG #URR8PPP up vote -1 down vote favorite Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point. (i.e $partial E neq emptyset $). ================= Here is what I tried. Consider $P_0=(x_1,x_2,dots,x_n)in E,P_1=(y_1,y_2,dots,y_n)notin E $. Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,dots,ty_n+(1-t)x_n) $, $0le tle 1$. $ t_0=supt $. And then I wanted to prove that $P_t_0in partial E$. A. If $P_t_0in E$. then $tneq 1$ otherwise $P_t_0=P_1$. And by definition ,$P_t notin E$ for $t_0 lt t leq 1 $. And choose $t_n$,such that $1gt t_ngt t_0$,$t_n to t_0$, which makes $P_t_n notin E$, but $P_t_n to P_t_0$.Then $P_t_o in partial E$. B. If $P_t_0notin E$. then $tneq 0$ otherwise $P_t_0=P_0$.And then choose $t_n$ such that $0lt t_nlt t_0$ , $t_n to t_0$ ,therefore $P_t_n to P_t_0$ and $P_t_n in E$. Hence $P_t_o in partial E$. Thus we have $partial E neq emptyset$. Am I correct? the construct of $P_t$ is a clue from my elder. What...