Prove that if sequence does not tend to infinity, then it has a bounded subsequence?

Question

I need to prove that that if sequence does not tend to infinity, then it has a bounded subsequence. I have tried to divide my problem into 2 subcases. 1) when my sequence is bounded, by Bolzano-Weierstrass I can prove that it has a bounded subsequence. 2) when my sequence is unbounded, it also has 2 subcases. I do contradiction and assume that there is no subsequence which is bounded. Either my subseqences tend to infinity or they do not have a limit. I'm stuck at this point. What would you suggest to do next? enter image description here

Answer 1

In my opinion it is useless to divide the problem in these two subcases because if sequence is bounded obviously it has bounded subsequence (it itself is its own sub-sequence).

I suggest a solution: let's assume that there is sequence $(a_n)$ which does not tend to infinity. It means that the following is not true:

$$\forall N > 0\, \exists n: \forall k > n \, |a_k| > N.$$

The opposite of the logical expression is $$\exists N > 0\, \forall n: \exists k > n \, |a_k| \leq N.$$ Let's construct bounded subsequence. We will fix the $N$ from expression and for the first element one can take $a_{k_1}$ such that $k_1 > 1,|a_{k_1}| \leq N$ (indeed it exists because one can put 1 instead of $n$ in expression $\exists N > 0\, \forall n: \exists k > n \, |a_k| \leq N$). For the next term we will get $a_{k_2}$ such that $k_2>k_1,a_{k_2} \leq N$. Continuing the process we will have bounded subsequence.