# Subsequential limit of $\mathbb Q$ in $\mathbb R$

Let $\{p_n\}$ be a sequence whose range is all of $\mathbb Q$. Prove that the set of subsequential limits of $\{p_n\}$ in $\mathbb R$ is all of $\mathbb R$

My theorm that if $\{p_{n_i}\}$ as a subsequence of $p_n$ converges, then it's limit is the subsequential limit of $p_n$, but isn't this telling me that $\mathbb Q$ converges to $\mathbb R$? I'm thinking this because $\{p_{n_i}\}$ is a subsequence of $\mathbb Q$ and so it's also a subsequence of $\mathbb R$?

• $\mathbb{Q}$ is not a sequence and does not "converge" to $\mathbb{R}$, which also is not a sequence. ($\mathbb{Q}$ is dense in $\mathbb{R}$, its closure is $\mathbb{R}$.) – BrianO Oct 26 '15 at 23:10

If you believe that every real number is the limit of a sequence of rational numbers, then for every $x\in \Bbb{R}$, pick such a sequence $\{q_n\}\subset \Bbb{Q}$ that converges to $x$. Now, since the range of $\{p_n\}$ is $\Bbb{Q}$, there exists for all $n$ an integer $k_n$ such that $p_{k_n}=q_n$. So $\{p_{k_n}\}_{n \in \Bbb{N}}$ converges to $x$.
• Sorta done :) Exercise for the reader (OP): The elements $(p_{k_n})_{n \in\mathbb{N}}$ will be in a different order than those of $(q_n)$. It remains to show that this isn't a problem. – BrianO Oct 26 '15 at 23:08
Let $r \in \Bbb R$. For all integer $k > 0$, choose $n_k$ such that $n_k > n_i$ for all $i < k$ and such that $r - {1\over k} < p_{n_k} < r + {1\over k}$. Since there are infinitely many rationals in any non-empty open interval, this is always possible. Then $p_{n_k}$ will be a subsequence that converges to $r$.