Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f(\xi)=0$?

share|improve this question
    
Related question –  Pedro Tamaroff Jul 31 '12 at 23:19

1 Answer 1

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 $L$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.