Why does every limited succession have a convergent subsuccession? Working on some real analysis homework, I faced this question:


*Let $(a_n)\ n ∈ N$ a real number succession. Prove that $(a_n)\ n ∈ N$ does not have a converging subsuccession if and only if $\lim_{n → +∞} |a_n| = + \infty $
I solved it being a sufficient condition, but I couldn't prove it being a necessary one. I came to the conclusion that any succession which has no limit but doesn't go to infinity is limited, but nothing beyond.
I asked my professor and he said that the result that every limited succession have a convergent subsuccession is a given, but didn't prove it.
How does one go about proving it?
 A: Let me rephrase this in a more standard way: you want to show that a bounded sequence in $\mathbb{R}$ has a convergent subsequence. This is a sketch of the proof (it should be a exercise to make this rigorous): say that $$|a_n|\leq R\quad \forall\:n\in\mathbb{N},$$ which means that $$a_n\in[-R,R]\quad \forall\:n\in\mathbb{N}.$$
Now, consider the half intervals $[-R,0]$ and $[0,R]$. One of them must have infinitely many terms of the sequence, right? Say that it's $[0,R]$ and consider $[0,\frac{R}{2}]$ and $[\frac{R}{2},R]$; again one of them has to have infinitely many terms of the sequence.
You can continue this process indefinitely and get a sequence of nested intervals whose length goes to $0$. Take $a$ to be the intersection of all this intervals. Then $x$ is an accumulation point of $\{a_n\}_n$, in other words a subsequence of $\{a_n\}_n$ converges to $a$.
Hope it helps.
PS: this is a classical result from real analysis called the Bolzano–Weierstrass theorem.
A: If $\lim\limits_{n\to\infty}|a_n| \neq \infty$, then it has a bounded subsequence: if it did not, then for any $C$, there would be some $n$ such that for all $m > n$, $|a_n| > C$, so $\lim\limits_{n\to\infty}|a_n| = \infty$, a contradiction. Choose some $C$ at which this fails.
We now proceed by "lion hunting":
Define $b_0 = -C$ and $c_0 = C$. We shall then construct sequences $(b_n)$ and $(c_n)$ such that there are always infinitely many terms of $a_n$ between them. We shall do this inductively as follows:
Let $m_n$ be the midpoint of $b_n$ and $c_n$. Then at least one of $[b_n, m_n]$ and $[m_n, c_n]$ contains infinitely many terms of $a_n$ (we know that there are infinitely many in $[b_0,c_0]$, and by our inductive hypothesis, in $[b_n, c_n]$). Choose $b_{n+1}$ and $c_{n+1}$ to be the endpoints of that interval.
Now, $(b_n)$ and $(c_n)$ are monotone and bounded. Define $b = \limsup b_n$. We shall show that $(b_n) \to b$.
For the former, note that, by definition of $b$, for any $\varepsilon > 0$, there is some $n$ such that $0 < b - b_n < \varepsilon$. But then, for any $m > n$, we have $0 < b - b_m < b - b_n < \varepsilon$. Thus, $(b_n) \to b$.
We note also that $b_n < c_n$ for all $n$, so $b$ lies in $[b_n, c_n]$ for all $n$.
Let $n_k$ be the smallest natural number such that $a_{n_k} \in [b_k, c_k]$. Then we have $|a_{n_k} - c| \leq |c_k - b_k| = 2^{k-1}C \to 0$, and so $(a_{n_k}) \to c$, and in particular, that $(a_{n_k})$ is a convergent subsequence of $(a_n)$.
A: $\lim_n |a_n|\neq \infty$ can be rewritten as $$\kappa:=\liminf_{n\to \infty}|a_n| \in [0,\infty).$$
Hence, there exists a subsequence $(|a_{n_k}|: k\ge 1)$ which is bounded, i.e., such that its image is contained in the compact set $[0,\kappa]$. Therefore by the Bolzano--Weierstrass theorem, there exists a sub-subsequence $(|a_{n_{k_t}}|: t\ge 1)$ which is convergent to some limit  $\ell \in [0,\kappa]$. It follows that there is an infinite set $T\subseteq \mathbf{N}$ such that $a_{n_{k_t}} \ge 0$ for all $t \in T$ or $a_{n_{k_t}} \le 0$ for all $t \in T$. Therefore $(a_{n_{k_t}}: t\in T)$ is convergent (to $\ell$ or to $-\ell$).
