Existence of an unbounded sequence I want to know why a sequence that is not bounded cannot have all of its subsequences converge to the same limit. According to Lebl, a bounded sequence is convergent and converges to x if and only if every convergent subsequence converges to x. I know that convergent subsequences must be bounded... But I'm having a hard time relating subsequences to their corresponding sequences in a way that implies something meaningful about the sequences. Can anyone give me hints?
 A: Here is an outline of a proof for you to fill out:
Theorem: Let $(x_n)$ be a sequence of real numbers. Let $x$ be a real number. Then $(x_n)$ converges to $x$ if and only if both of the following hold:


*

*$(x_n)$ is bounded. That is there are $a,b \in \Bbb R$ such that for all $n\in \Bbb N$, $a\le x_n \le b$.

*Every convergent subsequence of $(x_n)$ converges to $x$. That is, for each increasing sequence of integers $n_k$, if $(x_{n_k})$ converges, then $(x_{n_k})$ converges to $x$.


Lemma 1: If $(x_n)$ is a convergent sequence of real numbers, then $(x_n)$ is bounded.
Proof: Suppose $(x_n)$ converges to $x$. Then for any $\epsilon>0$ there is an $N>0$ such that …. In particular, letting $\epsilon = 1$ ….
Lemma 2: If $(x_n)$ is a sequence of real numbers converging to $x$, and $(x_{n_k})$ is any subsequence thereof, then $(x_{n_k})$ converges to $x$.
Lemma 3: If $(x_n)$ is a bounded sequence of real numbers, then $(x_n)$ has a convergent subsequence.
Lemma 4: If $(x_n)$ is a bounded sequence of real numbers that does not converge to $x$ then it must have a subsequence that converges in $(x,\to)$ or in $(\gets,x)$.
A: Let's assume a sequence isn't bounded. You can easily have convergent subsequences. For example, take the sequence $1, 0, -1, 0, 2, 0, -2, 0...$, which is unbounded and has exactly one sequence that is convergent.
However, you can't have all of the subsequences of an unbounded sequence be convergent. For one, every sequence has itself as a subsequence. So that's one that isn't convergent. If you only want proper subsequences, let $S_n, n \in N$ be the numbers in the main sequence and make your subsequence $S' = S_2, S_3, S_4...$ Since $S_1 \notin S'$, it's a proper subsequence that is also unbounded.
Finally, we can show that there must exist subsequences that are unbounded even if you remove an infinite amount of terms. Since $S$ is unbounded, for every number $a > 0$ there is some $n$ st $S_n > a$. So let $\alpha(x)$ be the smallest $n$ such that $S_n > x$. Then the subsequence $S' = S_{\alpha(1)}, S_{\alpha(3)}, S_{\alpha(5)} ...$ is a proper subsequence of S and is unbounded.
Note: To make sure it's actually a real subsequence you have to modify it to $\alpha(x)$ is the smallest n such that $S_n > x$ and that $\alpha(n) > \alpha(n-1)$. Otherwise if S was $101, 102, 103...$ we'd get $S' = 101,101,101...$ which is not a subsequence.
A: Suppose you have a sequence $(x_n)$ that is not bounded. Remember that a sequence is bounded if you have some positive number $M$ such that $|x_n|<M$ for all $n$. The negation of this statement is as follows. For all $M>0$, there is some $n$ such that $|x_n|>M$.
Now, for each integer $k$, choose such an integer $n_k$ such that $|x_{n_k}|>k$. and consider the subsequence $(x_{n_k})_{k\in\mathbb{Z}_+}$. Clearly, this subsequence cannot converge, because its absolute value becomes arbitrarily large eventually.
To see this formally, suppose that $(x_{n_k})$ converges to some real $c$. Then, for all $\varepsilon>0$, there must exist an integer $K$ such that for all $k>K$, we have $$|x_{n_k}-c|<\varepsilon.$$ But for such values of $k$, $$k<|x_{n_k}|=|x_{n_k}-c+c|\leq|x_{n_k}-c|+|c|<\varepsilon +|c|.$$ However, $k$ must exceed $\varepsilon+|c|$ for $k$ large enough, leading to a contradiction.
