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Find a sequence which has an infinite amount of limit points.

I was thinking about using the bijective pairing function $\langle\cdot,\cdot\rangle:\Bbb N\times\Bbb N\to\Bbb N,\langle x,y\rangle=\binom{x+y+1}{2}+x$ with $\pi_1(\langle x,y\rangle)=x$ and $\pi_2(\langle x,y\rangle)=y$ to describe the sequence

$$a_n=\frac{\pi_1(n)}{\pi_2(n)}.$$

In this case all numbers in $\Bbb Q$ are part of $a_n$, thus all numbers from $\Bbb R$ should be limit points due to $\Bbb Q$ being dense in $\Bbb R$.

Is this solution right or do you have an even easier solution?

Christian Ivicevic
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    Easier and more intuitive : http://math.stackexchange.com/questions/9319/sinn-subsequence-limits-set – T_O Mar 31 '14 at 11:35
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    You could find a sequence that takes a number on the unit circle, and rotates it by an irrational angle, meaning it will never return to its end point and eventually 'cover' the entire unit circle. – Pol van Hoften Mar 31 '14 at 11:37
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    @user45878 Be carefull there - speaking in radians, the full circle has angle $2\pi$, which is irrational. In fact, in radians, any rational angle will work. The condition you're looking for is that $\frac{\theta}{2\pi}$ isn't rational, or in other words that $\theta$ and $\pi$ are incommensurable. – fgp Mar 31 '14 at 11:50
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    Choose any enumeration of the rationals. Then each $x\in \mathbb{R}$ is a cluster point (limit point) of $(x_n)$. – Jose Antonio Mar 31 '14 at 20:58

4 Answers4

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Let $X_n$ be the largest prime factor of $n$. All primes numbers will be limit points of this sequence. As there are an infinite number of prime numbers, we get the desired sequence.

user137846
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Here’s yet another example: Let $\{\lambda_n\}_n$ be an enumeration of all rational numbers; we know this can be done in many different explicit ways. Then every real number is a limit point of this sequence.

Lubin
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I know the answer: { 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 1, 6, ..., n, n-1, ..., 2, 1, ...}

InDev
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A sequence accumulating on all $0\le x \le 1$ can be constructed as: First fix an intger $k$, for.example $k=5$. Then $$ x_n = n/k, \qquad 0 \le n \le k $$ Continue by $$ x_n = \frac{n-k}{2k} \quad k+1\le n \le 3k $$, and finally, for each $i$, $$ x_n = \frac{n-(2^i-1)k}{2^i k} ,\quad (2^i-1)k+1 \le n \le (2^{i+1}-1)k $$ This sequence will have an uncountable number of accumulation points.

kjetil b halvorsen
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