A sequence whose sub-sequence can converge to any value in $\mathbb R$

Question

In my university calc class, a problem on one of our assignments was

Is there a sequence $\{a_n\}$ of real numbers such that for all $α ∈ \mathbb R$, there is a subsequence $\{a_{n_i}\}$ such that $\lim_\limits{i→∞} a_{n_i} = α$? Prove your answer.

My friend came up with a sequence consisting of increasing lines, with each line starting at $-i$ and increasing by $1/i$ until i is reached. So for example:

$a_1 = -1$

$a_2 = 0$

$a_3 = 1$

which marks the end of the first line. Then $i$ is now $2$, so: $a_4 = -2$, $a_5 = -1.5, \ldots$ etc. infinitely.

Using the $\epsilon-\delta$ proof of a limit, we can easily show that we can make a subsequence converge value of $α$ by taking the closest term to $α$ in each "line" , which will eventually converge to $α$. For example if $α = \pi$

$$a_{n_1} = 1,\quad a_{n_2} = 2,\quad a_{n_3} = 3,\quad a_{n_4} = 3\quad \dots \text{ etc.}$$

Is there any other way to approach this problem? I never would have come up with an answer like this so I was wondering what other solutions might be possible.

Answer 1

Consider the biyection between $\mathbb{Q}$ and $\mathbb{N}$, namely $f:\mathbb{N}\to\mathbb{Q}$. This is a sequence and, by the density of $\mathbb{Q}$ in $\mathbb{R}$, for all $\alpha\in\mathbb{R}$ there exist a subsequence of $f$ such that converges to $\alpha$

Answer 2

That idea is good (although it could be expressed more clearly…).

You can also use the fact that $\mathbb Q$ is countable. Therefore, there is a sequence $(a_n)_{n\in\mathbb N}$ of rational numbers such that each rational number is equal to some $a_n$. But then, since $\mathbb Q$ is dense in $\mathbb R$ (that is, for each real number, there are rational numbers arbitrarly close to it), every real number is the limit of some subsequence of $(a_n)_{n\in\mathbb N}$.