Can someone give a concrete example of a sequence of reals that has countable infinite many cluster points ?
Example of a sequence with countable many cluster points
2 Answers
First fix some bijection $f : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$. For $n \in \mathbb{N}$ let $g(n)$ denote the first coordinate of $f(n)$ and let $h(n)$ denote the second corrdinate.
Then define a sequence $\{ x_n \}_{n=1}^\infty$ by $x_n = g(n) + 2^{-h(n)}.$ Then every natural number is a cluster point of this sequence.
For a more concrete example, consider the following sequence: $\begin{array}{c|c} n & x_n \\ \hline 1 & 1 + 2^{-1} \\ 2 & 1 + 2^{-2} \\ 3 & 2 + 2^{-1} \\ 4 & 1 + 2^{-3} \\ 5 & 2 + 2^{-2} \\ 6 & 3 + 2^{-1} \\ 7 & 1 + 2^{-4} \\ 8 & 2 + 2^{-3} \\ 9 & 3 + 2^{-2} \\ 10 & 4 + 2^{-1} \\ \vdots & \vdots \end{array} $
(I'm too lazy to give an exact formula at the moment, but I think the idea is clear.)
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1I happen to have played around with such bijections recently: One useful choice for the inverse of $f$ is $n = f^{-1}(k,l) = 2^k(2l+1) - 1.$ If $f(n) = (g(n),h(n))$ then \begin{align*} k &= g(n) = \lfloor \log_2(n+1)\rfloor \\ l & = h(n) = \frac{1}{2}\left(2^{\{\log_2(n+1)\}}-1\right).\end{align*} where $\{x\}$ denotes the fractional part of $x$. – 2012-09-05
-- For $\,n=1,3,5,..., 2n-1,...\,$ , define $\,a_n=1\,$
-- For $\,n=2,6,10,...,2+4(n-1),...\,$ , define $\,a_n=2\,$
-- For $\,n=4,12,20,..,4+8(n-1),...\,$ , define $\,a_n=3\,$
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-- For $\,n=2^k,2^k+2^{k+1},...,2^k+2^{k+1}(n-1),...\,$ , define $\,a_n=k+1$
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Now just take the sequence $\,\left\{a_n\right\}_{n=1}^\infty\,$