# Is there a sequence $(a_n)$ so that for every $r \in \mathbb{R}$, there is a subsequence of $(a_n)$ convergent to r?

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?

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

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) $$\square$$.

• @nicomezi Yes, thanks Jul 18, 2023 at 5:58
• 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. Jul 18, 2023 at 6:00
• @Arthur which part is not working for general topo case?
– Tim
Jul 18, 2023 at 6:27
• @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. Jul 18, 2023 at 9:05
• @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. Jul 18, 2023 at 11:10

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 $$r$$.