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Im unsure of an example Give an example of a sequence of real numbers with subsequences converging to every real number

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HINT: The rationals are a countable set, so they can be enumerated, and every real is the limit of a sequence of rationals.

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It's a bit of work, but you may show that taking the sequence $$a_n = n \bmod{2\pi}$$ has a limit point at every point in the interval $[-\pi, \pi]$, where the endpoints points are never actually attained attained (otherwise $\pi$ would be rational). So you may continuously map the interval $(-\pi,\pi)$ into $\mathbb{R}$ in you favorite way. Take, for instance, the map $f: (-\pi,\pi) \to \mathbb{R}$ sending $x \mapsto \tan(x)$. So now you have a sequence $a_n$ converging to every point $a \in (-\pi,\pi)$, then since tangent is continuous, $\tan(a_n) \to \tan(a)$, and $\tan$ is surjective onto $\mathbb{R}$, so you are done.

I realize that the first part, taking integers modulo $2\pi$ is not entirely necessary for the proof, since $\tan(x) = \tan(x + 2k\pi)$, but I think it's easier to break it into smaller problems.

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  • $\begingroup$ Or based on the opposite idea, $a_n = f(n\pi\mod 1)$ where $f(x) = \frac1x\sin\left(\frac1x\right)$. The modulo operation is necessary in this case, though. $\endgroup$ – user1551 Dec 12 '12 at 4:53
  • $\begingroup$ I think what you're referring to when you say $n\pi \bmod{1}$ is actually $n \pi - \lfloor n\pi \rfloor$ (the "floor" function). The way you have written is not well defined. i.e. $f(x)$ must be equal to $f(x + k)$ for any $k \in \mathbb{N}$, however $f(1) \neq f(2)$. $\endgroup$ – Andrew Maurer Dec 12 '12 at 5:02
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A related question that you can try:

Let $(a_k)_{k\in\mathbb{N}}$ be a real sequence such that $\lim_k a_k=0$, and set $s_n=\sum_{k=1}^na_k$. Then the set of subsequential limits of the sequence $(s_n)_{n\in\mathbb{N}}$ is connected.

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