a simple question about the density convergence of sequences Definition 1: A sequence $ \{x_n, n=1,2,3,...\}$  of points in a topological space $X$  converges to a point $x\in X$ in density if for any neighborhood $ V$  of $x$ in $X$, $x_n\in V$ but for a set of $n$ of density 0. We write $D-\lim x_n=x$.
Definition 2: Let S be a subset of N. The upper density of $S$, $D^{*}(S)$ is defined by $$D^{*}(S)=\limsup_{n\to \infty}  \#(S\cap [1,n])/n. $$ 
Question: A little property about the density convergence of a sequence is that, if $ \{n_k, k=1,2,3,...\}$ is a subset of positive upper density of $N$ and $D-\lim x_n=x$, then $D-\lim  x_{n_k}=x $ ,too.
This is a simple property. However, I can't come up with it. Help me ,please.
 A: Let $E \subset \mathbb N$ be the set corresponds to the subsequence $\{n_k: k\in \mathbb N\}$. The assumption is that $E$ has positive upper density $C$. 
Claim 1 We have 
$$\limsup_{k\to \infty} \frac{k}{n_k} \geq C . $$
Proof of Claim 1 We have 
$$\limsup_{n \to \infty} \frac{ E \cap \{1, \cdots, n\}}{n} \geq C$$ 
Note that if $n_k < n < n_{k+1}$, then 
$$ \frac{|E\cap \{1, \cdots, n\}|}{n} = \frac{|E\cap \{1, \cdots, n_k\}|}{n}< \frac{|E \cap \{1, \cdots ,n_k\}|}{n_k}$$
Thus we have 
$$\limsup_{k \to \infty} \frac{k}{n_k}= \limsup_{k\to \infty} \frac{|E\cap \{1, \cdots, n_k\}|}{n_k} \geq C$$
and that finishes the proof of claim 1. 
Claim 2 Let $S$ be a subset in $\mathbb N$ of density zero then $\tilde S = \{ k: n_k \in S\}$ is of density zero. 
Proof of Claim 2 Let $\epsilon >0$. Then there is $N \in \mathbb N$ so that 
$$\frac{|S \cap \{1, \cdots, n\}|}{n} < \epsilon$$
for all $n\geq N$. In particular, there is $K \in \mathbb N$ so that 
$$\frac{|S \cap \{1, \cdots, n_k\}|}{n_k} < \epsilon$$
for all $k \geq K$ (Just choose $K$ so that $n_K \geq N$). Note that $|\tilde S_k| \leq |S \cap \{1, \cdots, n_k\}|$, where $ \tilde S_k = \tilde S \cap \{1, \cdots, k\}$. Hence
$$\frac{|\tilde S \cap \{1, \cdots, k\}|}{n_k} < \epsilon$$
whenever $k\geq K$. Now by Claim 1 (Caution!), there is $K'$ so that $k/ n_k \geq C/2$ for all $k \geq K'$. Thus 
$$\frac{|\tilde S \cap \{1, \cdots, k\}|}{k} =  \frac{|\tilde S \cap \{1, \cdots, k\}|}{k} \frac{n_k}{k}\leq  \frac{|\tilde S \cap \{1, \cdots, k\}|}{n_k} \frac{2}{C}\leq \frac{2\epsilon}{C}$$
whenever $k\geq \max\{K, K'\}$. As $\epsilon >0$ is arbitrary, $\tilde S$ is of density zero and claim 2 is proved. 
Now we go back to your question. Let $V$ be an open set in $X$ containing $x$. Let $S$ be the subset so that $x_n \notin V$ whenever $n \in S$. By the $D$- convergence of $(x_n)$, $S$ is of density zero. Now let $(x_{n_k})$ be a subsequence of $(x_n)$. Write $\tilde S \subset \mathbb N$ so that $x_{n_k} \notin V$ whenever $k \in \tilde S$. Thus by claim 2, $\tilde S$ is also of density zero. Thus 
$$D- \lim_{k\to \infty} x_{n_k} = x . $$
Caution Note that as pointed out by David Chan, the answer is incomplete as 
$$\limsup_{k\to \infty} \frac{k}{n_k} \geq C$$ 
is not sufficient to show $\frac{k}{n_k} \geq \frac{C}{2}$ for large $k$. 
