How to prove the function $y = \sin x$ is not a closed function?

Question

I came across a question:

Suppose that $f(x) = \sin x$ is a function from $\mathbb R$ to $[-1,1]$. How do I prove the function $f(x) = \sin x$ is not a closed function?

By "closed function", I mean a function such that the image of any closed set is closed.

Answer 1

Closed function is a function such that image of every closed set is closed.

It is relatively easy to see that, for any $\varepsilon>0$, every $\varepsilon$-discrete subset of real line is closed. (A subset $A$ of a metric space $(X,d)$ is called $\varepsilon$-discrete if for any two distinct points $x,y\in A$ we have $d(x,y)\ge\varepsilon$. For subsets of real line, this condition means $|x-y|\ge\varepsilon$.)

Can you find a sequence $(x_n)$ with the following properties?

• $x_n\in(2n\pi,(2n+1)\pi)$ (which implies that $\{x_n; n\in\mathbb N\}$ is an $\varepsilon$-discrete subset for any $\varepsilon<\pi$)
• $\lim\limits_{n\to\infty} \sin x_n =y$ but $y\notin\{\sin x_n; n\in\mathbb N\}$

If $(x_n)$ fulfills the above properties, then $A=\{x_n; n\in\mathbb N\}$ is a closed set, but the image of this set is not closed.