Let $E$ be a compact set and let $f$ be continuous on $E$. Suppose there is a sequence {$x_n$}$_{n=1 \to \infty}$ of such points of E such that $\lvert f(x_n) \rvert < \frac{1}{n}$ for each $n$. Explain why there is a point y of E such that $f(y)=0$
So basically, I'm given that $f(x_n)$ is compact. Also since $E$ is compact, there is a point in $E$ (call it $y$) where $y$ is in $E$ and {$x_n$}$_{n=1 \to \infty}$ converges to $y$. (These are just my thoughts)
Intuitively I know that $\lvert f(x_n) \rvert < \frac{1}{n}$ and if $\exists y$ in $x_n$, where $f(y)=0$, but how do I show this ?
The sequence $\{x_n\}$ may not converge. Since $E$ is compact, $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to some element $y \in K$. Now apply continuity.