In the proof of existence of zero point, $f(x)$ is continuous in $[a,b]$, where $f(a)<0$ and $f(b)>0$.
It is shown on the proof process in the textbook that when we define a set $V$ as follows: $V=\{x |f(x)<0,x\in[a,b]\},$ so, there exists the supremum for $V$. Take $\xi=\sup V$.
Then I was confused with the following step:
take $x_{n}\in V (n=1,2,... \ )$, $x_{n}\rightarrow\xi$ (when $n\rightarrow \infty$) then
$f(\xi)=\lim_{n\rightarrow\infty}{f(x_{n})} \le0$
I know that $f(\xi)=\lim_{n\rightarrow\infty}{f(x_{n})}$ cause $f(x)$ in continuous in $[a,b]$
but why $f(\xi)=0$?