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I am looking for a sequence with infinitely many limit points. I don't want to use $\log,\sin,\cos$ etc.!

It's easy to find a sequence like above, e.g. $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,\dots$

But how can you prove the limit points? The problem I am having is the recursion or definiton of the sequence which I can't name exactly. But for a formal proof I need this.

So what's a sequence with infinitely many limit points without using $\log,\sin,\cos$ or any other special functions?

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You could let the sequence be your favorite bijection $\mathbb N\to\mathbb Q$. Then every real number is a limit point! –  Henning Makholm Nov 24 '13 at 5:21

2 Answers 2

up vote 1 down vote accepted

You can use something like :{ $1/n$}$\cup${$1+1/n$} $ \cup ....\cup$ {$k+1/n$} $\cup.... $

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What you have written is the set {$k+1/n:k\in\Bbb N$}. How does this define a sequence? –  John Bentin Nov 24 '13 at 14:06
    
I thought I'd just give a hint by giving a countable set with infinitely-many limit points. It seems I get chided for giving away too little as well as for giving away too much of the answer. Still, now the answer has been accepted, let me try to write it as a sequence. –  user99680 Nov 24 '13 at 20:01
    
But the countable set {$k+1/n:k\in \Bbb N$} doesn't have any limit points. Did you perhaps mean {$k+1/n:k,n\in \Bbb N$}? –  John Bentin Nov 24 '13 at 22:39
    
@John Bentin: precisely; re your other post, I will just state that the countable union of countables is countable, so that the union can be enumerated into a sequence. I don't know if this is too lazy; I know there are proofs that produce an actual sequence from this union. –  user99680 Nov 24 '13 at 23:08

Let $\{r_n\}$ be the set of all rationals. Then sequence $\{r_1, r_1, r_2, r_1, r_2, r_3, r_1, r_2, r_3, r_4,r_1,r_2,r_3,r_4,r_5,...\}$ has all rationals as limit points.

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