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Let be $(a_n)_{n\in\mathbb{N}}$ a bounded sequence of real numbers and we want to show that it contains a convergent subsequence.


My approach:

We know, by the least upper bound property, that the set $M:=\{a_n\mid n\in\mathbb{N}\}$ has a supremum $S$. As $S$ is the least upper bound of $M$, for each $\epsilon>0$ there exists a $a_n$ such that $S-\epsilon<a_n<S+\epsilon$.

Let be $\epsilon>0$ then there exists a $k_0\in\mathbb{N}$ such that $\frac{1}{k_0}<\epsilon$. We define a subsequence $\left( a_{n_k}\right)_{k\in\mathbb{N}}$ as follows:

There exists a $a_n$ which we set $a_{n_{k_0}}:=a_n$ such that

$$ S-\frac{1}{k_0}<a_{n_{k_0}}<S+\frac{1}{k_0}. $$ For all $k>k_0$ it follows that $$ S-\frac{1}{k_0}<S-\frac{1}{k}<a_{n_k}<S+\frac{1}{k}<S+\frac{1}{k_0}\\ \implies |a_{n_k}-S|<\frac{1}{k}<\epsilon. $$

Hence, $(a_n)_{n\in\mathbb{N}}$ is convergent.


My tutor said that it is flawed!? I have spent hours revising it and couldn't find any mistakes?

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    $\begingroup$ Unfortunately your tutor is right :P (tutor vs student 1-0). The problem on the construction you are making is in the indices. If you repeat again your argument you will see that you skipped rather fast the point of the construction of the subsequence. How can you pick $k_{n+1}>k_n$ such that $S\geq \alpha_{k_{n+1}}>S-1/k$? For example take the sequence $\alpha_n=-1$ for all $n\geq 2$ and $\alpha_1=1$. Then $S=1$ but there is no subsequence that converges to $1$. $\endgroup$ – dem0nakos Dec 12 '20 at 15:19
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The flaw is that the selected term may always be the same one. In particular if the difference of one term with the other ones is greater than a fix number.

Example. Take $\{a_n\}$ the always vanishing sequence except that $a_1 =1$. Then $S = 1$ but you won't be able to select a converging subsequence to $1$. Whatever $k \in \mathbb N$ you take, the only $a_n$ satisfying $ \vert a_n - 1 \vert \lt 1/k$ is $a_1$. Remember than for a subsequence $\{a_{n_k}\}$, $n_k$ has to be strictly increasing with $k$.

You can however build a subsequence converging to $\limsup\limits_{n} a_n$.

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Other comments and answers have told you where your errors are.

The existential problem with this question is the following: When you have seen just a finite number of the $a_n$'s you have no idea where a point of convergence could be. This means that you have to encompass the full sequence $(a_n)_{n\geq1}$ at each step of the process, in order to find a point of convergence.

We may assume $0\leq a_n\leq1$ for all $n$. Let $S$ be the set of all $x\in{\mathbb R}$ such that for infinitely many $n$ we have $a_n\geq x$. Then $0\in S$, and $S$ is bounded above by $1$. The set $S$ therefore has a supremum $\sigma\in{\mathbb R}$. I claim that there is a subsequence $k\mapsto a_{n_k}$ with $\lim_{k\to\infty} a_{n_k}=\sigma$.

Proof. Put $n_1:=1$, and assume that for a $k\geq1$ the "selection" $n_k$ has been constructed such that $|a_{n_k}-\sigma|\leq{1\over k}$. We then shall find an $n_{k+1}>n_k$ such that $$|a_{n_{k+1}}-\sigma|\leq{1\over k+1}\ .\tag{1}$$

By definition of the sup there is an $s\in S$ with $s>\sigma-{1\over k+1}$, and there are infinitely many $a_n\geq s$. On the other hand there are only finitely many $a_n>\sigma+{1\over k+1}$. It follows that there is an $n_{k+1}>n_k$ satisfying $(1)$.

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  • $\begingroup$ Would it be also correct to say that there are only finitely many $a_n$ such that $a_n>\sigma$? $\endgroup$ – Philipp Dec 14 '20 at 16:59
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    $\begingroup$ @Philipp: No. When $a_n={1\over n}$ $\,(n\geq1)$ then $S={\mathbb R}_{\leq0}$ and $\sigma=0$. There are infinitely many $a_n>\sigma$, but only finitely many $a_n>\sigma+\epsilon$ when $\epsilon>0$. $\endgroup$ – Christian Blatter Dec 14 '20 at 19:16

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