TITLE: Let $f\colon \mathbb R\to\mathbb R$ be a continuous function such that $f(i) = 0$
QUESTION [9 upvotes]: Let $f\colon \mathbb R\to\mathbb R$ be a continuous function such that $f(i) = 0$, for all $i\in \mathbb Z$ . Which of the following statement is true$?$
$(A)$ $\operatorname{Image}(f)$ is closed in $\mathbb R$.
$(B)$ $\operatorname{Image}(f)$ is open in $\mathbb R$.
$(C)$ $f$ is uniformly continuous.
$(D)$ None of the above.
If we take $f(x) = 0$, for all $x\in \mathbb R$, then $\operatorname{Image}(f)$ is a singleton set. Also we know that every singleton set is closed in $\mathbb R$. So option $B$  is wrong.
For option $(C)$, I take $f\colon \mathbb R\to\mathbb R$ such that $f(x) = x \sin\pi x $, this function $f$ is not uniformly continuous. So option $C$ is also wrong.
Now my problem is that the answer of this question is option $D$ but i am not able to get any function to discard option $A$. So please help me.
Thanks you.

REPLY [8 votes]: Take $$f(x)=\frac{\sin\pi x}{1+e^{-x}}.$$ Then show $\mathrm{Im}(f)=(-1,1)$.
Or:
$$g(x)=\sin\pi x \arctan x$$
Then $\mathrm{Im}(g)=\left(-\frac \pi 2,\frac \pi 2\right).$
Basically, this is because $x\mapsto \frac{1}{1+e^{-x}}$ and $x\mapsto\arctan x$ are increasing functions with images $(0,1)$ and $(-1,1)$ respectively.