[proofplan] Each of the eight properties is proved independently using the $\varepsilon$-definition of convergence: given $\varepsilon > 0$, produce an index $N \in \mathbb{N}$ beyond which the desired inequality holds. The principal tools are the triangle inequality (for uniqueness, sums, and products), the boundedness of convergent sequences (for the product rule), the reverse triangle inequality (for the reciprocal rule), and proof by contradiction (for bound preservation). Part (viii) is reduced to part (vii) by negation. [/proofplan] [step:Prove uniqueness of limits via the triangle inequality] Assume $a_n \to a$ and $a_n \to b$. Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that \begin{align*} |a_n - a| < \frac{\varepsilon}{2} \quad \text{for all } n \geq N_1 \end{align*} and \begin{align*} |a_n - b| < \frac{\varepsilon}{2} \quad \text{for all } n \geq N_2. \end{align*} Let $N = \max(N_1, N_2)$. For any $n \geq N$, the triangle inequality gives \begin{align*} |a - b| \leq |a - a_n| + |a_n - b| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Since $\varepsilon > 0$ is arbitrary, $|a - b| = 0$, hence $a = b$. [/step] [step:Show subsequences inherit the convergence threshold] Assume $a_n \to a$ and let $(n_j)_{j=1}^{\infty}$ be a strictly increasing sequence of natural numbers. Given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $|a_n - a| < \varepsilon$ for all $n \geq N$. Since $(n_j)$ is strictly increasing, an induction argument shows $n_j \geq j$ for all $j \in \mathbb{N}$. Therefore, for $j \geq N$ we have $n_j \geq j \geq N$, so \begin{align*} |a_{n_j} - a| < \varepsilon. \end{align*} This establishes $a_{n_j} \to a$. [/step] [step:Verify constant sequences converge] Let $a_n = c$ for all $n \in \mathbb{N}$. Given $\varepsilon > 0$, for any $n \geq 1$: \begin{align*} |a_n - c| = |c - c| = 0 < \varepsilon. \end{align*} Therefore $a_n \to c$ (with $N = 1$ serving for every $\varepsilon$). [/step] [step:Establish the sum rule by splitting $\varepsilon$] Assume $a_n \to a$ and $b_n \to b$. Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that \begin{align*} |a_n - a| < \frac{\varepsilon}{2} \quad \text{for all } n \geq N_1 \end{align*} and \begin{align*} |b_n - b| < \frac{\varepsilon}{2} \quad \text{for all } n \geq N_2. \end{align*} Let $N = \max(N_1, N_2)$. For $n \geq N$, the triangle inequality yields \begin{align*} |(a_n + b_n) - (a + b)| &\leq |a_n - a| + |b_n - b| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $a_n + b_n \to a + b$. [/step] [step:Prove the product rule using boundedness of convergent sequences] Assume $a_n \to a$ and $b_n \to b$. Since every convergent sequence is bounded, there exists $M > 0$ such that $|a_n| \leq M$ and $|b| \leq M$ for all $n$. Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that \begin{align*} |a_n - a| < \frac{\varepsilon}{2M} \quad \text{for all } n \geq N_1 \end{align*} and \begin{align*} |b_n - b| < \frac{\varepsilon}{2M} \quad \text{for all } n \geq N_2. \end{align*} Let $N = \max(N_1, N_2)$. For $n \geq N$, the add-and-subtract identity $a_n b_n - ab = a_n(b_n - b) + b(a_n - a)$ gives \begin{align*} |a_n b_n - ab| &= |a_n(b_n - b) + b(a_n - a)| \\ &\leq |a_n||b_n - b| + |b||a_n - a| \\ &\leq M \cdot \frac{\varepsilon}{2M} + M \cdot \frac{\varepsilon}{2M} = \varepsilon. \end{align*} Therefore $a_n b_n \to ab$. [/step] [step:Derive the reciprocal rule via a lower bound on $|a_n|$] Assume $a_n \to a$, $a_n \neq 0$ for all $n$, and $a \neq 0$. The key preliminary step is to establish a lower bound for $|a_n|$ that holds for all sufficiently large $n$. Since $|a| > 0$, applying the definition of convergence with $\varepsilon_0 = \frac{|a|}{2}$ yields an index $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$: \begin{align*} |a_n - a| < \frac{|a|}{2}. \end{align*} The reverse triangle inequality then gives \begin{align*} |a_n| \geq |a| - |a - a_n| > |a| - \frac{|a|}{2} = \frac{|a|}{2}. \end{align*} Now, given $\varepsilon > 0$, there exists $N_1 \in \mathbb{N}$ such that for all $n \geq N_1$: \begin{align*} |a_n - a| < \frac{|a|^2 \varepsilon}{2}. \end{align*} Let $N = \max(N_0, N_1)$. For $n \geq N$: \begin{align*} \left| \frac{1}{a_n} - \frac{1}{a} \right| &= \left| \frac{a - a_n}{a \cdot a_n} \right| \\ &= \frac{|a - a_n|}{|a| \cdot |a_n|} \\ &< \frac{|a - a_n|}{|a| \cdot \frac{|a|}{2}} \\ &= \frac{2|a - a_n|}{|a|^2} \\ &< \frac{2}{|a|^2} \cdot \frac{|a|^2 \varepsilon}{2} = \varepsilon. \end{align*} Therefore $\frac{1}{a_n} \to \frac{1}{a}$. [/step] [step:Prove upper bound preservation by contradiction] Assume $(a_n)_{n=1}^{\infty}$ is a real sequence with $a_n \leq A$ for all $n$ and $a_n \to a$. Define $t_n = A - a_n \geq 0$. By the sum rule (applied to the constant sequence $A$ and the sequence $-a_n$), we have $t_n \to A - a$. Suppose for contradiction that $a > A$. Then $\delta := a - A > 0$, and since $t_n \to A - a = -\delta$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$: \begin{align*} |t_n - (A - a)| < \frac{\delta}{2}. \end{align*} This implies \begin{align*} t_n < (A - a) + \frac{\delta}{2} = -\delta + \frac{\delta}{2} = -\frac{\delta}{2} < 0, \end{align*} contradicting $t_n \geq 0$. Therefore $a \leq A$. [/step] [step:Reduce lower bound preservation to the upper bound case] Assume $(a_n)_{n=1}^{\infty}$ is a real sequence with $A \leq a_n$ for all $n$ and $a_n \to a$. Define $b_n = -a_n$ and $B = -A$. Then $b_n \leq B$ for all $n$, and by the sum rule (applied to the constant sequence $0$ and $-a_n$, or directly from the product rule with $-1$), we have $b_n \to -a$. By the upper bound preservation result of the preceding step, $-a \leq B = -A$, which gives $A \leq a$. [/step]