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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.

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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.)

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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.

$$y<\frac{k\pi}2\implies \left(k-\frac12\right)\pi\sin\left(\left(k-\frac12\right)\pi\right)<y<\left(k+\frac12\right)\sin\left(\left(k+\frac12\right)\pi\right)$$

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