# Proof that $x\sin(x)$ has infinite accumulation points

I have to find a sequence with infinite many accumulation points and intuitively I thought about $$\sin(x)$$ - since it is periodic, it has points from its codomain that get repeated infinite many times. However, not infinite many points since all the y values that come from a sine function are bounded in the interval [-1,1]. But what about $$xsin(x)$$? The definition of accumuation point is: let $$(a_n$$) be a sequence of real numbers. The number a is said to be an accumulation point of $$(a_n)$$ if there exists a subsequence $$(an_k)$$ such that $$\lim\limits_{k\to\infty} an_k=a$$. If my intuition is correct from the plots it looks like each point get repeated infinite many times, together with the growth of the x value (in both directions). However I also need to come up with a formal proof to back up (or destroy) my intuition and here is where I got stuck. I thought about trying to describe a subsequence by exploiting the periodical behaviour, like I can say that 0 is an accumulation point of $$sin(x)$$ by choosing the subsequence $$sin (\pi k),k\in\ \Bbb{Z}$$. Am I on the wrong path or does it make some sense? Any feedback would be greatly appreciated.

I believe you may have some confusion about how to construct your example. You want a sequence, but $$\sin x$$ and $$x\sin x$$ are both functions. I think I understand the gist of your idea, but it will be tricky to manufacture sequences out of these functions that do the job.
As a suggestion, try thinking about the rational numbers $$\mathbf Q$$. Since they form a countably infinite set, choose an enumeration of $$\mathbf Q$$, that is, a function $$\{1,2,3,\dots\}\to\mathbf Q$$, and let us denote it as $$\{q_n\}_{n=1}^\infty$$. To show that this has infinitely many accumulation points, think about why there is a subsequence $$\{q_{n_{k_1}}\}_{k_1=1}^\infty$$ such that $$\lim_{k_1\to\infty}q_{n_{k_1}}\to 1$$, a subsequence $$\{q_{n_{k_2}}\}_{k_2=1}^\infty$$ such that $$\lim_{k_2\to\infty}q_{n_{k_2}}\to 2$$, and so on. (Hint: density.)
Every real value $$y$$ is an accumulation point of $$x\sin x$$ because the equation
$$y=x\sin x$$ has infinitely many solutions. Indeed, the RHS alternates between $$-x$$ and $$x$$ with period $$2\pi$$, and as soon as $$x>y$$, $$y$$ is crossed twice by period.
$$y<\frac{k\pi}2\implies \left(k-\frac12\right)\pi\sin\left(\left(k-\frac12\right)\pi\right)