0
$\begingroup$

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 $\mathbb R$?

$\endgroup$
  • $\begingroup$ $\mathbb{Q}$ 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 $\mathbb{R}$.) $\endgroup$ – BrianO Oct 26 '15 at 23:10
3
$\begingroup$

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 $x$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – BrianO Oct 26 '15 at 23:08
  • $\begingroup$ Indeed, thanks for noting that. I guess that it is left to the reader :) $\endgroup$ – Nitrogen Oct 26 '15 at 23:10
  • $\begingroup$ I updated my comment with that suggestion :) $\endgroup$ – BrianO Oct 26 '15 at 23:13
1
$\begingroup$

Too late to the game, but just for reference, here is an approach that avoids the difficulty that BrianO has pointed out:

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 $r$.

$\endgroup$
  • $\begingroup$ No such thing as too late, I always appreciate someone taking the time to answer! $\endgroup$ – user181928 Oct 28 '15 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.