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Proving that if $f$ is continuous on an interval $I$, any $a,b\in I,a<b$ and for any $y$,where $f(a) < y < f(b)$, there exist a $x\in (a,b)$ s.t $f(x)=y$.

I have seen a prove using the sequence but i want to prove it with my understanding of the definition of continuity so i decided to prove it with contradiction. Assume there is a $y$ which have no $x\in (a,b)$ s.t.$f(x)=y$ and i wanna prove that $f$ is then not continuous but I'am not sure how to prove it with contradiction. The definition i learnt about the continuous function on a set $X$ is :$\forall x,y\in X,\forall \epsilon>0,\exists\delta>0$ s.t.$d(x,y)<\delta\implies d(f(x),f(y)) < \epsilon$.

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up vote 5 down vote accepted

Hint: Try the following approach: Let $A = \{x \in I: f(x) < y\}$. This set is nonempty since $a \in I$ and $f(a) < y$ by assumption. Let $\alpha = \sup A$ and show that $\alpha$ has the desired properties.

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Yup, this can prove the theorem but i wanna stick to the definition of continuous function so i prefer prove by contradiction as above mentioned. – Mathematics Jan 5 '13 at 15:02
I think with 'desired properties' anonymous meant that $f$ is not continuous at $\alpha$. By the definition you want to use this means $\exists\epsilon>0:~\forall \delta>0 \exists x\text{ with }d(x,\alpha)<\delta\text{ and }d(f(x),f(\alpha))\geq \epsilon$ – Michalis Jan 5 '13 at 15:07
@Michalis How you get the $\epsilon$ then? Is it just simply any $\epsilon<|y-f(\alpha)|$? – Mathematics Jan 5 '13 at 15:45
@Mathematics: Yes indeed! – Michalis Jan 5 '13 at 15:52

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