Why is $(1,1,...)$ an extreme point of a closed unit ball in $c$?

Question

Why is $(1,1,...)$ an extreme point of a closed unit ball in $c$ (the space of all convergent sequences with sup norm)?

Suppose on the contrary that $(1,1,...)= t (x_n)+(1-t) (y_n)$ for some $t$ in $(0,1)$ and $(x_n)\ne (y_n)\in c$. Then, $tx_n+(1-t)y_n=1$ for all $n$, so $t x+(1-t)y=1$, where $x_n\to x, y_n\to y$.

How do I get a contradiction from here to conclude that $(1,1,...)$ is indeed an extreme point?

Answer 1

$1 =tx_n+(1-t)y_n \leq t(1)+(1-t)(1)=1$. If $x_n<1$ or $y_n<1$ then we get $1 =tx_n+(1-t)y_n < t(1)+(1-t)(1)=1$ or $1<1$ which is a contradiction. Hence, the only possibility is $x_n=y_n=1$ for all $n$.

As pointed out by Anne Bauval the conclusion holds even for complex sequences. [If $|c|\leq 1$ and $\Re c=1$ then $\Im c=0$ necessarily, so $c=1$].