[The Order Limit Theorem] Assume \(\lim a_n = a\) and \(\lim b_n = b\). Then,
  1. If \(a_n \geq 0\) for all \(n \in \mathbb{N}\), then \(a \geq 0\);
  2. If \(a_n \leq b_n\) for all \(n \in \mathbb{N}\), then \(a \leq b\);
  3. If there exists \(c \in \mathbb{R}\) for which \(c \leq b_n\) for all \(n \in \mathbb{N}\), then \(c \leq b\). Similarly, if \(a_n \leq c\) for all \(n \in \mathbb{N}\), then \(a \leq c\)

Definitions of sequences and convergence: here, Definitions of subsequences: here.
For the “show the limit” template: here.

Proof (i)

Assume for the sake of contradiction that \(a < 0\). Since \(\lim a_n = a\), then for \(\epsilon > 0\), there exist some \(n \geq N\) such that \(|a_n - a| < \epsilon\). Choose \(\epsilon = |a|\). Then, we must have an \(n \geq N\) such that

$$ \begin{align*} |a_n - a| < |a| \\ |a_n - a| < |a| \\ |a_n| - |a| < |a|. \\ |a_n| < 0. \\ \end{align*} $$

But this contradicts the assumption that \(a_n \geq 0\) for all \(n \in \mathbb{N}\). Therefore, we must have \(a \geq 0\). \(\blacksquare\)

Proof (ii)

Write \(a_n \leq b_n\) as \(b_n - a_n \geq 0\). Since we have both \(\lim a_n = a\) and \(\lim b_n = b\), then by the algebraic limit theorem, This limit will converge to \(b - a\). By part (i), since \(b_n - a_n \geq 0\), then \(b-a \geq 0\) and so \(a \leq b\) as required. \(\blacksquare\)

Proof (ii) [Alternative]

Suppose for the sake of contradiction that \(b < a\). Then, let \(\epsilon = \frac{|a-b|}{2}\). Then

$$ \begin{align*} (a - \epsilon, a + \epsilon) \cap (b - \epsilon, b + \epsilon) = \emptyset \end{align*} $$

Now, since we’re given that \(\lim a_n = a\), then there exists an \(N_1 \in \mathbb{N}\) such that when \(n \geq N_1\), we have

$$ \begin{align*} |a_n - a| < \frac{|a-b|}{2} \text { or } a_n \in (a - \epsilon, a + \epsilon) \end{align*} $$

Similarly, we’re given that \(\lim b_n = b\), then there exists an \(N_2 \in \mathbb{N}\) such that when \(n \geq N_2\), we have

$$ \begin{align*} |b_n - b| < \frac{|a-b|}{2} \quad \text{ or } b_n \in (b - \epsilon, b + \epsilon) \end{align*} $$

Now, let \(N = \max(N_1, N_2)\). When \(n \geq N\),

$$ \begin{align*} a_n \in (a - \epsilon, a + \epsilon) \text { and } b_n \in (b - \epsilon, b + \epsilon) \end{align*} $$

which are disjoint internvals. But this implies that \(a_n > b_n\). This is a contradiction since the original assumption is that \(a_n \leq b_n\). Therefore, we must have \(a \leq b\). \(\blacksquare\)

Proof (iii)

TODO

References