Continuous real valued function that is zero on integers

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(i)=0 \;\; \forall i\in\mathbb{Z}.$ Which of the following statements is always true?

A) The image of $f$ is closed in $\mathbb{R}$

B) The image of $f$ is open in $\mathbb{R}$

C) $f$ is uniformly continuous

D) none of above

Since we can take $f$ to be zero function, option B is wrong. What about the others?

Answer 1

Hint: $f(x)=h(x)$Sin$(\pi x)$ satisfies $f(i)= 0$ $ \forall i \in \bf Z$,Now can you find suitable $h$ to contadict options $a$ and $c$ ?

Answer 2

For $n\in Z$ : For $x\in [n,n+1/2]$ let $f(x)=-(1-2^{-|n|})(x-n)(n+1/2-x).$ For $x\in [n+1/2,n+1]$ let $f(x)=|\sin (4\pi n(x-n-1/2))|$. The image of $f$ is $ (-1/4,1]$ which is neither open nor closed. And $f$ is not uniformly continuous because $f'(n+3/4)=4\pi |n| $, and $\{4\pi |n| :n\in Z\}$ is an unbounded set.