Let $(a_n)$ be a sequence of real numbers that is unbounded above. Show that $\exists$ a subsequence $(a_{n_k})_{k \ge 1}$ such that $\lim_{k \rightarrow \infty} a_{n_k} = + \infty$.

Working so far:

Since $(a_n)$ is not bounded above, this means that there exists no $M \in \mathbb{R}$ such that $a_n \le M$ for all $n \in \mathbb{N}$. In other words, there exist infinitely many $n$'s such that $a_n > M$ for any real $M$. I will prove by construction that there exists a strictly increasing subsequence $(a_{n_k})_{k \ge 1}$ of $(a_n)$ that diverges to $+\infty$.

For the first term of the subsequence, begin by picking $n_1$ to be the first index such that $a_{n_1} > 1$. This is possible because by assumption, $(a_n)$ is unbounded. For $k \ge 2$, pick $n_k > n_{k-1}$ to be the first index such that $a_{n_k} > k$ and $a_{n_k} > a_{n_{k-1}}$. I will use a proof by contradiction to show that we can always find such an index $n_k$, $k \ge 2$. If $n_k$, $k \ge 2$, does not exist, then either $a_n \le k$ or $a_n \le a_{n_{k-1}}$ for all $n \ge n_k$. But this contradicts the fact that $(a_n)$ is unbounded above. Hence, such an index $n_k$, $k \ge 2$, always exists.

Now I will show that the subsequence $(a_{n_k})_{k \ge 1}$ defined above diverges to $+\infty$, that is, $\lim_{k \rightarrow \infty} a_{n_k} = +\infty$. To do this, we need to show that for all $H > 0$, there exists an $N \in \mathbb{N}$ such that whenever $n_k \ge N$, then $a_{n_k} \ge H$.


1) Is my argument to construct the subsequence correct? I thought I might've needed to use induction, but is the contradiction statement correct?

2) I know this may seem obvious, but how exactly do I use the definition of diverging to $+\infty$ to finish my proof? That is, how do I pick $N$?


So to summarize, the above construction produces a subsequence $(a_{n_k})_{k \ge 1}$ such that for all $n_k < n_{k+1}$, we have $a_{n_k} < a_{n_{k+1}}$ and $a_{n_k} > k$. To show that this subsequence diverges, just pick $k$ large enough so that $k \ge H$ and then set $N = n_k$.


construction part is correct and nice. In the construction of required sequence, you have picked $n_k$ which satisfies so and so condition. Use the same index $n_k$ to prove it diverges.

  • $\begingroup$ Thanks, please see my edit. Basically we can just pick $k$ large enough so that $k \ge H$ and set $N = n_k$, right? $\endgroup$ – user40333 Apr 19 '14 at 9:09
  • $\begingroup$ ya. it is correct. $\endgroup$ – GA316 Apr 19 '14 at 9:42

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