Question about unbounded, real-valued function on closed interval.

Question

Is the following statement true? If $f:[a,b]\to [0,\infty)$ is unbounded, then for each $M>0$, there exists $c\in[a,b]$ and $\delta>0$ such that $f(x)>M$ whenever $\mid$x-c$\mid<\delta$. It seems like this should be the case, but I am having difficulty proving it. Obviously, if $f$ is unbounded, then for any $M>0$, there exists $c\in[a,b]$ where $f(c)>M$. I want to know if the same can be said for $f(x)$ when $x$ very close to $c$. I'm trying to use this as part of a proof involving Riemann Integrability and boundedness. Any pointers in the right direction (whether it's true of false, for starters), would be much appreciated.

Answer 1

You can define a function on $[0,1]$ by $f(x)=n$ when $x=\frac{1}{n}$ $(n \in \mathbb{N})$, $f(x)=0$ otherwise. It is clearly not bounded, but it equals something which is not zero only in isolated points and not in their neighborhoods. So your statement is not true.