Exercise 2.5.2

Question

Decide whether the following propositions are true or false, providing a short justification for each conclusion.
  \begin{enumerate}[label=(\alph*)] 
  \item If every proper subsequence of $\left(x_{n}\right)$ converges, then $\left(x_{n}\right)$ converges as well.
  \item If $\left(x_{n}\right)$ contains a divergent subsequence, then $\left(x_{n}\right)$ diverges.
  \item If $\left(x_{n}\right)$ is bounded and diverges, then there exist two subsequences of $\left(x_{n}\right)$ that converge to different limits.
  \item If $\left(x_{n}\right)$ is monotone and contains a convergent subsequence, then $\left(x_{n}\right)$ converges.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item True, removing the first term gives us the proper subsequence $(x_2, x_3, \dots)$ which converges. This implies $(x_1, x_2, \dots)$ also converges to the same value, since discarding the first term doesn't change the limit behavior of a sequence.
  \item True, as this is the contrapositive of "if a sequence converges then all subsequences converge (to the same limit)"
  \item True, since $x_n$ is bounded $\lim \sup x_n$ and $\lim \inf x_n$ both converge. And since $x_n$ diverges Exercise ef{ex:lim_sup} tells us $\lim \sup x_n 
e \lim \inf x_n$.
  \item True, The subsequence $(x_{n_k})$ converges meaning it is bounded $|x_{n_k}| \le M$. Suppose $(x_n)$ is increasing, then $x_n$ is bounded since picking $k$ so that $n_k > n$ we have $x_n \le x_{n_k} \le M$. A similar argument applies if $x_n$ is decreasing, therefore $x_n$ is monotonic bounded and so must converge.
   \end{enumerate}

Exercise 2.5.2

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Exercise 2.5.2

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