Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous, $f(0)>0$, and $f(1)=0$. Prove that there is a number $x_0 \in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x < x_0$; that is, there is a smallest point in the interval $[0,1]$ at which the function $f$ attains the value $0$.
Let $R=\{x\in(0,1]|f(x)=0\}$, $R$ is bounded below and is not empty since $f(1)=0$ where $1\in R$. So there exists an element $x_0$ in $(0,1]$ be the infimum of $R$. To yield a contradiction, assume that $x_0$ is not a element of $R$, then for all $n$, there exists a sequence $x_n$ in $R$ such that $x_0<x_n<x_0+1/n$. Thus $x_n$ converges to $x_0$ and gives us $f(x_n)=f(x_0)$ since $f(x)$ is continuous. But since for all n $x_n$ in $R$, so we have $f(x_n)=0=f(x_0)$ which contradicts with $x_0$ is not an element of $R$. Therefore, $x_0$ is the minimum element in $R$. To show $f(x)>0$ for $0\leq x<x_0$, we will apply contradiction. To yield a contradiction, we assume that there exists a number $x_1\in[0,x_0)$ such that $f(x_1)<0$. Since $f(0)>0$, now apply the Intermediate Value Theorem, there exists a number $x_2$ in $[0,x_0)$ such that $f(x_2)=0$, but this contradicts with $x_0$ is the minimum number. Therefore, there exists a number $x_0\in(0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\leq x<x_0$. $\square$
Can someone give me a hit or suggestion to write a proof without using the Intermediate Value Theorem and write a direct proof ? Thanks
