QUESTION THREE (20 MARKS) a) Define an \( \epsilon \)-neighbornood of a set. (2 marks) b) Show that arbitrary union of open subsets of \( \mathbb{R} \) is open. (5 marks) c) Given the sequence \( \left(\frac{n^{2}+n}{2 n}\right) \) i) Write down the first five terms of the sequence (3 marks) ii) State its monotonicity. (1 mark) iii) Determine its boundedness. (3 marks) d) Suppose that \( \left(x_{n}\right),\left(y_{n}\right) \) and \( \left(z_{n}\right) \) are sequences such that \( x_{n} \leq y_{n} \leq z_{n}, \forall n \in \mathbb{N} \). If \( \lim _{n \rightarrow \infty} x_{n}=l=\lim _{n \rightarrow \infty} z_{n} \), then show that \( \lim _{n \rightarrow \infty} y_{n}=l \). (6 marks)

### QUESTION THREE #### a) Define an \( \epsilon \)-neighborhood of a set. *(2 marks)* **Definition:** For a set \( A \subseteq \mathbb{R} \) and a positive real number \( \epsilon > 0 \), an **\( \epsilon \)-neighborhood** of \( A \) is the set of all points in \( \mathbb{R} \) that are within a distance \( \epsilon \) from at least one point in \( A \). Formally, \[ N_{\epsilon}(A) = \{\, x \in \mathbb{R} \mid \exists\, a \in A \text{ such that } |x - a| < \epsilon \,\} \] This means every point \( x \) in \( N_{\epsilon}(A) \) lies within the open interval \( (a - \epsilon, a + \epsilon) \) for some \( a \in A \). --- #### b) Show that arbitrary union of open subsets of \( \mathbb{R} \) is open. *(5 marks)* **Proof:** Let \( \mathcal{U} = \bigcup_{i \in I} U_i \), where each \( U_i \) is an open subset of \( \mathbb{R} \) and \( I \) is an arbitrary index set. To show that \( \mathcal{U} \) is open, take any point \( x \in \mathcal{U} \). By definition of union, there exists at least one \( U_j \) such that \( x \in U_j \). Since \( U_j \) is open, there exists an \( \epsilon > 0 \) such that the open interval \( (x - \epsilon, x + \epsilon) \subseteq U_j \). But \( U_j \subseteq \mathcal{U} \), hence \( (x - \epsilon, x + \epsilon) \subseteq \mathcal{U} \). Since every point \( x \in \mathcal{U} \) has an \( \epsilon \)-neighborhood entirely contained within \( \mathcal{U} \), it follows that \( \mathcal{U} \) is open. --- #### c) Given the sequence \( \left(\frac{n^{2}+n}{2 n}\right) \) ##### i) Write down the first five terms of the sequence. *(3 marks)* **Solution:** First, simplify the general term: \[ a_n = \frac{n^2 + n}{2n} = \frac{n(n + 1)}{2n} = \frac{n + 1}{2} \] Now, compute the first five terms: \[ \begin{align*} a_1 &= \frac{1 + 1}{2} = 1 \\ a_2 &= \frac{2 + 1}{2} = 1.5 \\ a_3 &= \frac{3 + 1}{2} = 2 \\ a_4 &= \frac{4 + 1}{2} = 2.5 \\ a_5 &= \frac{5 + 1}{2} = 3 \\ \end{align*} \] **First Five Terms:** \( 1, \ 1.5, \ 2, \ 2.5, \ 3 \) ##### ii) State its monotonicity. *(1 mark)* **Answer:** The sequence \( \left( \frac{n + 1}{2} \right) \) is **monotonically increasing** because each subsequent term is larger than the previous one. ##### iii) Determine its boundedness. *(3 marks)* **Analysis:** The general term of the sequence is: \[ a_n = \frac{n + 1}{2} \] As \( n \) increases, \( a_n \) grows without bound: \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n + 1}{2} = \infty \] **Conclusion:** - **Bounded Below:** The sequence is bounded below by \( 1 \) (the first term). - **Unbounded Above:** The sequence increases without bound. **Overall:** The sequence is **unbounded**. --- #### d) Suppose that \( \left(x_{n}\right), \left(y_{n}\right) \), and \( \left(z_{n}\right) \) are sequences such that \( x_{n} \leq y_{n} \leq z_{n} \) for all \( n \in \mathbb{N} \). If \( \lim_{n \to \infty} x_{n} = l = \lim_{n \to \infty} z_{n} \), then show that \( \lim_{n \to \infty} y_{n} = l \). *(6 marks)* **Proof:** We apply the **Squeeze Theorem**. 1. **Given:** \[ x_n \leq y_n \leq z_n \quad \text{for all } n \in \mathbb{N} \] \[ \lim_{n \to \infty} x_n = l \quad \text{and} \quad \lim_{n \to \infty} z_n = l \] 2. **For any \( \epsilon > 0 \),** since \( \lim_{n \to \infty} x_n = l \) and \( \lim_{n \to \infty} z_n = l \), there exists \( N \in \mathbb{N} \) such that for all \( n \geq N \): \[ |x_n - l| < \epsilon \quad \text{and} \quad |z_n - l| < \epsilon \] This implies: \[ l - \epsilon < x_n \leq y_n \leq z_n < l + \epsilon \] 3. **Thus, for all \( n \geq N \):** \[ |y_n - l| < \epsilon \] 4. **Conclusion:** By the definition of the limit, since \( |y_n - l| < \epsilon \) for all \( n \geq N \), it follows that: \[ \lim_{n \to \infty} y_n = l \] **Therefore, \( \lim_{n \to \infty} y_n = l \).**