91Ó°ÊÓ

Let's assume that a given sequence \(\left(a_{n}\right)\) has no convergent subsequence. We want to show that the absolute value of its terms tends to infinity, i.e., \(\left|a_{n}\right| \rightarrow \infty\). If \(\left(a_{n}\right)\) doesn't have a convergent subsequence, then for every subsequence \(\left(a_{n_k}\right)\), we can find an open interval, \(I\), such that the number of the elements of \(I\) in the subsequence is infinite. In particular, for any real number \(M\), we refine our interval and choose \(I = (-M, M)\). This implies that there are infinitely many elements in \(a_{n_k}\) out of \((-M, M)\). Now, take those elements that are not in the interval \((-M, M)\). Since there are infinitely many elements, there is an index \(n_0\) such that \(i > n_0\) implies \(\left|a_i\right| > M\). That means, for any \(M > 0\), there exists an index \(n_0\) such that for all \(n > n_0\), \(\left|a_{n}\right| > M\). Therefore, \(\left|a_{n}\right| \rightarrow \infty\), as we wanted to prove.

Now, let's assume that the absolute value of a sequence's terms tends to infinity, i.e., \(\left|a_{n}\right| \rightarrow \infty\). We want to show that the sequence has no convergent subsequence. Suppose, for contradiction, that there is a convergent subsequence \(\left(a_{n_k}\right)\). Let \(L\) be the limit of this subsequence, i.e., \(\lim_{k \rightarrow \infty} \left(a_{n_k}\right) = L\). Since the absolute value of \(a_n\) tends to infinity, for any \(M > 0\), there exists an index \(n_0\) such that for all \(n > n_0\), \(\left|a_{n}\right| > M\). Consider the subsequence \(\left(a_{n_k}\right)\). Since it converges to \(L\), for any \(\epsilon > 0\), there exists an index \(k_0\) such that for all \(k > k_0\), \(\left|a_{n_k} - L\right| < \epsilon\). Now, let's choose \(\epsilon = 1\). Then, for all \(k > k_0\), we have \(\left|a_{n_k} - L\right| < 1\). But since \(\left|a_{n}\right| \rightarrow \infty\), we can find a large enough \(M > \max \{|L|+1, L+1\}\), and by the definition of the limit, for all \(n > n_0\), \(\left|a_{n}\right| > M\). However, for some \(k' > k_0\), we have \(n_{k'} > n_0\). This leads to a contradiction since: $$\left|a_{n_{k'}} - L\right| < 1 \implies \left| a_{n_{k'}} \right| < \left| L \right| + 1 < M \leq \left| a_{n_{k'}} \right|$$ This contradiction implies that there is no convergent subsequence. Our proof is complete. In conclusion, we have shown that a sequence \(\left(a_{n}\right)\) in \(\mathbb{R}\) has no convergent subsequence if and only if \(\left|a_{n}\right| \rightarrow \infty\).