# How to prove there exists a subsequence of $x_n$ that converges to $x$ in the following case?

Let $x_n$ be a sequence in $\mathbb{R}$ and let $x\in \mathbb{R}$, how to prove that if $\forall\epsilon>0, \forall k\in \mathbb{N}, \exists n\ge k$ such that $|x_n-x|<\epsilon$, then there exists a subsequence of $x_n$ that converges to $x$?

One proof from online goes like this: Let $\epsilon$ and $k$ be given. Choose $n_1$ such that $|x_{n_1} - x| <1$. Inductively, for each $k=2,3,..$ choose $n_k$ such that $n_k>n_{k-1}$ and $|x_{n_{k}}-x|<\frac1k$. Then $x_{n_k}$ is a subsequence of $x_n$ and it converges by squeeze theorem.

I am quite confused about the proof. How are we allowed to choose $n_1$? How do you assume that such subsequence with starting point $n_1$ even exists? Also, how do you do the inductive part so that $|x_{n_{k}}-x|<\frac1k$? How does squeeze theorem work in this case?

• $n_1$ exists if $\epsilon=k=1$ – hamam_Abdallah May 21 '17 at 23:04

By assumption, there is an $n_{1}\geq 1$ such that $|x_{n_{1}}-x|<1$ (Here you are using $\epsilon=1$ and $k=1$).

Now, using and $\epsilon=1/2$ and $k=n_{1}+1$, there is an $n_{2}\geq n_{1}+1$ such that $|x_{n_{2}}-x|<1/2$.

Next, using $\epsilon=1/3$ and $k=n_{2}+1$, there is an $n_{3}\geq n_{2}+1$ such that $|x_{n_{3}}-x|<1/3$.

Continuing in this way, at the $m$th stage choosing $\epsilon=1/m$ and $k=n_{m-1}+1$, you get a an increasing sequence

$$n_{1}<n_{2}<n_{3}<\cdots$$ So, we may form the subsequence $x_{n_{1}},x_{n_{2}},x_{n_{3}},\ldots$ of $(x_{n})$.

We claim that this subsequence converges to $x$. Let $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $1/N<\epsilon$. Then, for all $m\geq N$, we have $$|x_{n_{m}}-x|<\frac{1}{m}\leq\frac{1}{N}<\epsilon.$$ Thus, the subsequence $(x_{n_{m}})_{m\in\mathbb{N}}$ converges to $x$.

$1.$ We are allowed to choose such an $n_1$ because the problem tells us so. If you unravel the meaning of the logical operators in the question statement, you will observe that for any legal choices of $\epsilon$ and $k$ we can always find a term in our sequence $\{x_n\}$ which satisfies the condition of being in a punctured epsilon neighborhood about $x$. It is up to us what to call such a term. If the authors of the proof you found have decided to call it $x_{n_1}$ it is likely to increase readability;

$2.$ How we assume a starting point exists is covered in $(1)$ above;

$3.$ We are not really assuming the ensuing sub-sequence to exist - that would be circular. Instead we are producing an "algorithm" of sorts, to construct it. A formal work-out of this method can be found in ervx's answer. If you want some intuition as to why this might work, consider the following. The question tells us that if we choose $x_{n_m}$ and $\epsilon = m$, where $m \in \mathbb{N}$, we can find some term $x_{n_{p}}$, where $n_{p} > n_m$, which satisfies $|x_{n_p} - x| < 1/m$. We can then play this game again, using a larger value for $m$, denote this as $m'$, to produce a new term, $x_{n_{p'}}$, which is an even better approximation to $x$. Playing this game for arbitrarily large values of $m$, and each time making sure to pluck the new $x_{n_p}$ term and placing it into a new sequence $S$, we find that the sequence $S$ constitutes a sub-sequence of $\{x_n\}$ which is converging to the value $x$;

$4.$ The intuition behind why the squeeze theorem works in this case is something as follows. Since the sub-sequence we have constructed has the nice property that each $n_k$ term is in a $k^{-1}$ neighborhood of $x$, if we let $k$ grow arbitrarily large we know that our terms will be included in very very small neighborhood of the limit value $x$ (i.e. for $k = 10^{10}$, our $x_{n_k}$ will differ by less than $10^{-10}$ away from the value of $x$). Increasing $k$ will only make this error smaller, indicating that we are "tending" towards our limit value of $x$.