Exercise 2.3.10

Question

Consider the following list of conjectures. Provide a short proof for those that are true and a counterexample for any that are false.
  \begin{enumerate}[label=(\alph*)] 
  \item If $\lim \left(a_{n}-b_{n}\right)=0$, then $\lim a_{n}=\lim b_{n}$.
  \item If $\left(b_{n}\right) \rightarrow b$, then $\left|b_{n}\right| \rightarrow|b|$.
  \item If $\left(a_{n}\right) \rightarrow a$ and $\left(b_{n}-a_{n}\right) \rightarrow 0$, then $\left(b_{n}\right) \rightarrow a$.
  \item If $\left(a_{n}\right) \rightarrow 0$ and $\left|b_{n}-b\right| \leq a_{n}$ for all $n \in \mathbf{N}$, then $\left(b_{n}\right) \rightarrow b$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item False, consider $a_n = n$ and $b_n = -n$.
  \item True since if $|b_n - b| < \epsilon$ then $||b_n| - |b|| \le |b_n - b| < \epsilon$ by Exercise 1.2.6 (d).
  \item True by ALT since $\lim (b_n - a_n) + \lim a_n = \lim b_n = a$.
  \item True, since $0 \le |b_n - b| \le a_n$ we have $a_n \ge 0$. Let $\epsilon > 0$ and pick $N$ such that $a_n < \epsilon$ for all $n \ge N$. Therefor
    $$
    |b_n - b| \le a_n < \epsilon
    $$
    Proving $(b_n) 	o b$.
   \end{enumerate}

Exercise 2.3.10

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Exercise 2.3.10

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