Suppose a function $f:\mathbb R \to \mathbb R$ is continuous with $f(0)=1$. Show that if there is a positive number $x_0$ for which $f(x_0)=0$, then there is a smallest positive number $p$ for which $f(p)=0$.
I think we want to find the infimum of the set $\{x |x>0, f(x)=0\}$ but I am unable to proceed further. Any help is appreciated. Thanks.
Your suggestion is absolutely correct. The set $\{x\in\mathbb R^+:f(x)=0\}\subset \mathbb R$ is obviously bounded below and therefore (by completeness of $\mathbb R$ has an infimum, say $x_0$.
We now prove $f(x_0)=0$. This follows from the series criterion for continuity. In more detail: Since $x_0$ is the infimum of $\{x\in\mathbb R^+:f(x)=0\}$ there exists a series $(x_n)\subset\{x\in\mathbb R^+:f(x)=0\} $ with $x_n\to x_0$. By continuity we then have $$ f(x_0)=\lim f(x_n)=\lim 0 = 0. $$ Further, the construction of $x_0$ as infimum yields $x_0\ge 0$. Since $f(x_0)=1$, $x_0\ne 0$ is obvious and $x_0>0$ follows immediately.