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Recently I came across the following result :

Let $A \subseteq \mathbb{R}^n$ be a connected set and $f : A \to \mathbb{R}$ a continuous function. If there are two vectors $u$ and $v$ such as $f(u)<0$ and $f(v)>0$ then there exists $w \in A$ such that $f(w)=0$.

Unfortunately the author did not prove this nor did he give any references. How can this be proven provided the fact that there are no constraints on $f$ ?

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This cannot possibly be true without some assumption on $f$. For instance, let $A = \mathbb{R}^d$ take $f(t) = -1$, for all $t \neq v$ and $f(v) = 1$, for any $u$ and $v \in \mathbb{R}^d$. – anonymous Dec 28 '12 at 21:28
Perhaps the result you "came across" assumes that $f$ is continuous. – GEdgar Dec 28 '12 at 21:32
You might just be right @GEdgar I will edit the question. – user54549 Dec 28 '12 at 21:35
up vote 3 down vote accepted

Assuming $f$ is at least continuous, the result is straightforward. Since $A$ is connected, so is $f(A)$. Thus it's an interval, and so the result follows since $f(u) < 0$, and $f(v) > 0$, we have that $(f(u), f(v)) \subset f(A)$.

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+1 Nice and simple – DonAntonio Dec 28 '12 at 21:43
How can $(f(u), f(v)) \subset f(A)$ if $(f(u), f(v)) \subset \mathbb{R}^2$ and $f(A) \subset \mathbb{R}$? – Mihai Bişog Jan 13 '13 at 15:21

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