9
$\begingroup$

My guess is that there is no such sequence and prove it by contradiction.

My attempt is as follows: Let $A_r$ denote the set containing all terms of such subsequence for each $r \in \mathbb{R}$. Since each subsequence that converges to $r$ contains at least one distinct term, then the union $\bigcup_{r\in \mathbb{R}}A_r$ has uncountably many elements. However, the sequences cannot have uncountably many terms.

Is there anything wrong with my attempt? Or a better solution to go?

$\endgroup$
3
  • 1
    $\begingroup$ Enumerate the rational numbers. $\endgroup$
    – Chad K
    Jul 18, 2023 at 5:43
  • 3
    $\begingroup$ Your error is in "each subsequence ... contains at least one distinct term". There is no reason to believe that. $\endgroup$
    – Ingix
    Jul 18, 2023 at 9:50
  • $\begingroup$ The union is a subset of the sequence, so at most infinitely many terms. $\endgroup$
    – Tim
    Jul 18, 2023 at 10:06

2 Answers 2

8
$\begingroup$

Answer to your question.
Let $\mathbb{Q}$ be the set of rational numbers in $\mathbb{R}$. This set is dense in $\mathbb{R}$ and it is countable. Let $q\colon\mathbb{N}\to \mathbb{Q}$ be a bijection and let $a_n$ be the $n$-th element of the sequence $0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,\dots$ (you start from $0$ and add $1$, every time you reach a number that hasn't been reached before, you reset back to $0$). Then $(q_{a_n})_{n\in\mathbb{N}}$ is a sequence that has an accumulation point at each element in $\mathbb{R}$. Indeed, let $r\in\mathbb{R}$, and let $\varepsilon >0$ and $N\in\mathbb{N}$. Since $\mathbb{Q}$ is dense, there is a $p\in\mathbb{Q}$ such that $|r-p|<\varepsilon$. Let $m\in\mathbb{N}$ be such that $q_m=p$ (remember that $q\colon\mathbb{N}\to\mathbb{Q}$ is bijective) and let $n>N$ be such that $a_n=m$. Then $q_{a_n}=q_m=p$ and therefore $|r-q_{a_n}|<\varepsilon$. In conclusion, $\forall \varepsilon>0,\forall N\in\mathbb{N},\exists n>N,\; |r-q_{a_n}|<\varepsilon$ and we have that there must be a subsequence of $(q_{a_n})_{n\in\mathbb{N}}$ that converges to $r$.

But make it topological.
This property characterizes separability of topological spaces:
Proposition. Let $(X,\mathscr{T})$ be a topological space. $(X,\mathscr{T})$ is separable if and only if there exists a $\varphi\in X^\mathbb{N}$ such that for every $x\in X$, there exists a subsequence of $\varphi$ that converges to $x$.
Proof. Let $\varphi$ be one such sequence, then $\varphi(\mathbb{N})$ is countable and every element of $X$ can be approached by some sequence in $\varphi(\mathbb{N})$, that is, $X\subseteq\overline{\varphi(\mathbb{N})}$. We conclude $\overline{\varphi(\mathbb{N})}=X$ and $X$ is separable. On the other hand, if $X$ is separable, let $Q$ be a countably dense set, biject it by $q\colon \mathbb{N}\to Q$ then $(q_{a_n})_{n\in\mathbb{N}}$ is the desired sequence ($a_n$ defined just like in the previous section) $\square$.

$\endgroup$
8
  • $\begingroup$ @nicomezi Yes, thanks $\endgroup$ Jul 18, 2023 at 5:58
  • 1
    $\begingroup$ For the special case of $\Bbb Q\subseteq \Bbb R$, there is no need to use $a_n$, the sequence $q_n$ alone does the job just fine. This wouldn't work for the general topological case, though. $\endgroup$
    – Arthur
    Jul 18, 2023 at 6:00
  • 1
    $\begingroup$ @Arthur which part is not working for general topo case? $\endgroup$
    – Tim
    Jul 18, 2023 at 6:27
  • 1
    $\begingroup$ @Tim If you take an arbitrary separable topological space, pick a dense countable subset, and enumerate it, then that enumeration as a sequence doesn't necessarily have a subsequence that converges to all points. Consider, for instance, a countably infinite discrete space, where the resulting sequence would have no convergent subsequences at all. You need to infinitely repeat each element as outlined in this answer. $\endgroup$
    – Arthur
    Jul 18, 2023 at 9:05
  • 1
    $\begingroup$ @Tim Not quite. Every open neighborhood of $r$ contains some $p$. That's the definition of separable. It's just that some times there are only finitely many possible $p$ (or even just one), and once you've used them, you're done, and can't get close to $r$ ever again. So you have to do the infinite repetition $0,0,1,0,1,2,\ldots$ thing to multiply the finitely many into infinitely many so that you actually get an infinite subsequence. $\endgroup$
    – Arthur
    Jul 18, 2023 at 11:10
1
$\begingroup$

I see it as a triangle expanding to the right where the n-th column has the rationals $(k/n)$ for $k=-n^2$ to $n^2$, so these are from $-n$ to $n$ spaced $1/n$ apart.

For any real $r$, once $n > |r|$, the n-th column has an entry within $1/n$ of $r$, so by making $n$ large enough, there is an entry within $1/n$ of $r$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .