# Find a sequence $(u_n)$ such that: $\forall l\in \mathbb R$, there exists a subsequence of $(u_n)$ converging to $l$.

This was a question I had in my first analysis exam. I am wondering if my proof works. The proof I had in mind during the exam works as follows:

Let $(u_n)$ be the sequence given by the Cantor diagonalization for $\mathbb Q$ ($u(1)=\frac{1}{1}$, $u(2)=\frac{2}{1}$, $u(3)=\frac{1}{2}$ $u(4)=\frac{1}{3}$ and so on).

We notice that $\mathbb R= \bigcup_{a\in\mathbb Z} [a,a+1]$ and so can divide $\mathbb R$ into subintervals of the form $[a,a+1]$. Without loss of generality, we consider such a subinterval $[a,a+1]$.

We consider then $(u_{f(n)})$, a subsequence of $(u_n)$ such that $\forall n\in\mathbb N, a\leq u_{f(n)}\leq a+1.$ As $a,a+1\in \mathbb Q$ it is clear that $\varlimsup_{n\to \infty} u_{f(n)}=a+1$ $\varliminf_{n\to \infty} u_{f(n)}=a$. We define a subsequence $u_{h(n)}$ of $u_{f(n)}$ such that $\lim_{n \to \infty}u_{h(n+1)}-u_{h(n)}=0$ in the following way:

$$u_{h(1)}=a, u_{h(2)}=a+1, u_{h(3)}=a+\frac{1}{2}, u_{h(4)}=a , u_{h(5)}=a+\frac{1}{3}, u_{h(6)}=a+\frac{2}{3}$$ and so on (we can guarantee these are all in $u_{f(n)}$ as they are all rational and within the given interval). In general, we note that $\forall k$ such that $h\big(\frac{n(n+1)}{2} \big)< k\leq h\big(\frac{(n+1)(n+2)}{2}\big)$, $u_k-u_{k-1}=\frac{1}{n+1}$. It is clear then that the desired property holds.

Let $l'\in [a,a+1]$. We consider the subsequence $(u_{g(n)})$ of $(u_{h(n)})$ such that

$$\forall n\in \mathbb N, (u_{g(n)}\leq l' \leq u_{g(n+1)}) \vee (u_{g(n)}\geq l' \geq u_{g(n+1)}).$$ We rearrange each expression to

$$(0\leq l'-u_{g(n)}\leq u_{g(n+1)}-u_{g(n)}) \vee (0\geq l'-u_{g(n)}\geq u_{g(n+1)}-u_{g(n)}).$$

As $(u_{g(n)})$ is a subsequence of $(u_{h(n)})$ and $\lim_{n \to \infty}u_{h(n+1)}-u_{h(n)}=0$, it follows that $\lim_{n \to \infty}u_{g(n+1)}-u_{g(n)}=0.$ Passing the limit of the expressions above we find that

$$\big(0\leq l'-\lim_{n\to \infty} u_{g(n)} \leq \lim_{n\to \infty}(u_{g(n+1)}-u_{g(n)})\big) \vee \big(0\geq l'-\lim_{n\to \infty} u_{g(n)} \geq \lim_{n\to \infty}(u_{g(n+1)}-u_{g(n)})\big).$$

For each disjunct, we find that $l'-\lim_{n\to \infty}u_{g(n)}=0.$ Consequently $$l'=\lim_{n\to \infty} u_{g(n)}.$$

Therefore $\forall l'\in [a,a+1], \exists$ subsequence of $(u_n)$, denoted $(u_{g(n)})$, such that $l'=\lim_{n\to \infty} u_{g(n)}.$

As $\mathbb R=\bigcup _{a\in \mathbb Z} [a,a+1]$, we can thus conclude that $\forall l\in \mathbb R, \exists$ subsequence of $(u_n)$, denoted $(u_{g^*(n)})$, such that $l=\lim_{n\to \infty} u_{g^*(n)}.$

I asked the TA about this question and was told this would receive partial credit, so where exactly does the proof go astray? If any part of this were to go astray, it seems to me my definition of $u_{h(n)}$ may be somewhat unclear, though I am not sure. I also apologize for any unclear typesetting.

• You lost me at "it is clear $\limsup = a+1$." If you only know $a \leq u_{f(n)} \leq a+1$ then $\limsup \leq a+1$, not necessarily equal. – Michael Oct 27 '17 at 5:21
• I would just take any listing of the rationals $\{u_n\}_{n=1}^{\infty}$. Then for any real number $l$, there must be an infinite number of rationals in the list that are as close as you like to $l$. So choose the first $u_i$ that is a distance at most 1 from $l$, then choose $u_j$ with $j>i$ that is a distance at most $1/2$ from $l$, and so on. – Michael Oct 27 '17 at 5:23
• Why does anything have to go astray? Can't the marker take off marks for making the solution so long and complicated? – bof Oct 27 '17 at 8:16
• @bof Absolutely fair enough. I really didn't mean to question the grader's ability to grade as they are instructed/want to. – mnewman Oct 27 '17 at 12:51
• @Michael Thank you, Michael. So I would need to further justify that claim. The idea to me was that $sup_{n\geq 1} {u_{f(n)}}=a+1$ as $a+1\in \mathbb Q$. And so it seemed like it must be true in the limit that $limsup_{n \to \infty} u_{f(n)}=a+1.$ Would that have been sufficient justification? Admittedly manipulation of supremum and infimum has been my weakest point in this course. – mnewman Oct 27 '17 at 12:58

Rephrasing Michael's idea a bit:

Let $(x_n)_{n\in \mathbb{N}}$ be the sequence of all rational numbers.

Since $\mathbb{Q}$ is countable, a sequence

$(x_n)_{n \in \mathbb{N}}$ exists.

Let $l\in \mathbb{R}.$

Construct a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ that converges to l.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$:

There are $x_n$ in every neighbourhood of $l.$

Choose $x_{n_k}$ such that $|x_{n_k} -l| \lt 1/k.$

The subsequence $(x_{n_k})_{k \in \mathbb{N}}$ converges

to $l.$