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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 $f(\xi)=0$?

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    Related [question](http://math.stackexchange.com/questions/144316/proof-of-bolzanos-theorem)2012-07-31

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Well, you know $f(\xi)\leq 0$. Assume it is smaller, then by continuity there is a whole neighbourhood of $\xi$ where $f<0$. Therefore $\xi$ is not the supremum of $V$, which is a contradiction.

Edit: If you also don't understand why $f(\xi)\leq 0$, note that it follows from the more general fact that for a converging series where all (but finitely many) elements are smaller than some given $L$, then the limit is smaller or equal to $L$.

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    yeah i've made it clear.thanks2012-07-31