Convergent subsequences and natural density of sets of indices Does there exist a divergent sequence $(x_n)$ of real numbers such that for every d in $(0,1]$ and every $S \subseteq \mathbb{N}$ of natural density $d$, the subsequence $(x_i\,:\,i \notin S)$ is convergent? Every convergent sequence obviously satisfies this property, but I can't come up with a divergent example, nor a proof that such a sequence doesn't exist.
 A: No.  
Given a divergent sequence $(x_n)$, the sequence is either unbounded or it has at least $2$ limit points (Bolzano–Weierstrass), so each case will be considered.
First suppose it is unbounded.  Choose $n_1>1$ such that $|x_{n_1}|>1$, then $n_2>\max\{4,n_1\}$ such that $|x_{n_1}|>2$, and generally given $n_1<n_2<\ldots<n_{k-1}$, $n_k$ is chosen such that $n_k>\max\{k^2,n_{k-1}\}$ and $|x_{n_k}|>k$.  Define $S=\mathbb N\setminus\{n_1,n_2,n_3,\ldots\}$.  Then $S$ has density $1$ but $(x_{n_k})$ diverges.
If instead we suppose $(x_n)$ has two limit points $a\neq b$, the previous case can be modified by choosing $n_k>\max\{k^2,n_{k-1}\}$ as before, but now with the condition that $|x_{n_k}-a|<\frac1k$ if $k$ is odd, $|x_{n_k}-b|<\frac1k$ if $k$ is even.  Take $S$ the same. 
A: Another kind of answer:
Take $m\geq 3$, $S_0=\{km, k\geq 0\}$, $S_1=\{km+1, k\geq 0\}$, and $T=\mathbb{N}-(S_0\cup S_1)$. Then the density of $S_k$, $k=0,1$, is $1/m$, so the two sequences $(x_n, n\not \in S_0)$ and $(x_n, n\not \in S_1)$ are convergent, say to $L_0$ and $L_1$. As $(x_n, n\in T$) is a commun subsequence of both these sequences, we get $L_0=L_1$, and we see easily that $x_n$ is convergent. 
