Let $f\colon \mathbb R\to\mathbb R$ be a continuous function such that $f(i) = 0$

Question

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.

Answer 1

Take a function whose graph is a triangle connecting $(0, 0)$ to $(1/2, 1/2)$ to $(1, 0)$.

Then connect $(1, 0)$ to $(3/2, 3/4)$ to $(2, 0)$.

Then connect $(2, 0)$ to $(5/2, 7/8)$ to $(3, 0)$.

Continue in this manner, and the range is $[0, 1)$, which is not closed in $\mathbb{R}$.

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Moral of the story: The image $f([i, i + 1])$ is closed for each $i$. But a countable union of closed sets isn't always closed.

Answer 2

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.