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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 – Pedro Tamaroff Jul 31 '12 at 23:19
up vote 1 down vote accepted

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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thanks, i am confused that why $f(\xi)\leq0$.your explanation helps a lot. additionally, i wonder whether the 'general fact' that explain why$f(\xi)\leq 0$ is a inference of monotone convergence theory?i think i've forget former theory and should have a review – Francis King Jul 31 '12 at 12:23
Francis, this fact can be proven directly from the definition. Just assume you have a given converging sequence of the descirbed type and assume further that the limit is bigger than $L$, can you deduce a contradiction? – Simon Markett Jul 31 '12 at 12:42
@ Simon Markett .For a converging series where all (but finitely many) elements are smaller than some given L, we assume that $\lim_{n\rightarrow \infty}{f(x)}=R\ (R>L\ )$.then $\forall n>N$ $x_{n}>R-\epsilon$.Furthemore, cause $\epsilon$ can be infinitely small,$R-\epsilon>L$,so $\forall n>N\ (infinite \ ) $ $x_{n}>L$ .a contradiction to assumption: 'but finitely many'.Is this right way to deduce? – Francis King Jul 31 '12 at 13:47
Exactly, except that you probably meant $x_n$ when you wrote $f(x)$. I think you have everything together now for a full proof. – Simon Markett Jul 31 '12 at 13:59
yeah i've made it clear.thanks – Francis King Jul 31 '12 at 14:37

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