PPSC Lecturer Mathematics: Real Analysis and Topology Solved MCQs with Detailed Explanations

Dear Candidates,

As a lecturer preparing for the Punjab Public Service Commission (PPSC) or Federal Public Service Commission (FPSC) examinations, mastering Real Analysis and General Topology is crucial. These two subjects form the core of pure mathematics and account for a significant percentage of the MCQs in the PPSC written test.

Part 1: Key Theoretical Foundations

1. Real Analysis Fundamentals

• Completeness Axiom: Every non-empty subset $S \subset \mathbb{R}$ that is bounded above has a least upper bound (supremum) in $\mathbb{R}$, denoted as $\sup(S)$.
• Metric Space: A set $X$ equipped with a metric $d: X \times X \to \mathbb{R}_{\ge 0}$ satisfying non-negativity, identity of indiscernibles, symmetry, and the triangle inequality: $$d(x, z) \le d(x, y) + d(y, z) \quad \forall x, y, z \in X$$
• Uniform Continuity: A function $f: E \to \mathbb{R}$ is uniformly continuous on $E \subset \mathbb{R}$ if for every $\epsilon > 0$, there exists $\delta > 0$ (depending only on $\epsilon$) such that: $$|x - y| < \delta \implies |f(x) - f(y)| < \epsilon \quad \forall x, y \in E$$
• Heine-Borel Theorem: A subset $K \subset \mathbb{R}^n$ is compact if and only if it is closed and bounded.
• Sequence of Functions: A sequence $(f_n)$ converges uniformly to $f$ on $E$ if: $$\lim_{n \to \infty} \sup_{x \in E} |f_n(x) - f(x)| = 0$$

2. General Topology Fundamentals

• Topological Space: A pair $(X, \tau)$ where $\tau \subseteq \mathcal{P}(X)$ contains $\emptyset$ and $X$, and is closed under arbitrary unions and finite intersections.
• Closure and Interior: The interior $\text{Int}(A)$ is the largest open set contained in $A$. The closure $\overline{A}$ is the smallest closed set containing $A$. The boundary is defined as $\partial A = \overline{A} \setminus \text{Int}(A)$.
• Separation Axioms:
  • $T_0$ (Kolmogorov): For any distinct points, at least one has an open neighborhood not containing the other.
  • $T_1$ (Fréchet): Every singleton set $\{x\}$ is closed. Distinct points have neighborhoods not containing each other.
  • $T_2$ (Hausdorff): Distinct points $x \neq y$ possess disjoint open neighborhoods $U, V$ such that $U \cap V = \emptyset$.
• Compactness: A topological space $X$ is compact if every open cover of $X$ has a finite subcover.
• Connectedness: A space $X$ is connected if it cannot be written as the union of two disjoint non-empty open sets.

Part 2: Fully Solved Analytical Examples

Example 1: Uniform Continuity Analysis

Problem: Show whether $f(x) = x^2$ is uniformly continuous on $(0, 1)$ and on $[0, \infty)$.

Solution:

Case 1: On $(0, 1)$
The derivative $f'(x) = 2x$ is bounded on $(0, 1)$, since $|f'(x)| = |2x| < 2$. By the Mean Value Theorem: $$|f(x) - f(y)| = |f'(\c)| |x - y| \le 2 |x - y|$$ For any $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$. If $|x - y| < \delta$, then $|f(x) - f(y)| < 2 \cdot \frac{\epsilon}{2} = \epsilon$. Thus, $f(x) = x^2$ is uniformly continuous on $(0, 1)$.

Case 2: On $[0, \infty)$
Take $x_n = n + \frac{1}{n}$ and $y_n = n$. Then: $$|x_n - y_n| = \frac{1}{n} \to 0 \quad \text{as } n \to \infty$$ However, $$|f(x_n) - f(y_n)| = \left| \left(n + \frac{1}{n}\right)^2 - n^2 \right| = \left| n^2 + 2 + \frac{1}{n^2} - n^2 \right| = 2 + \frac{1}{n^2} \ge 2$$ Since $|f(x_n) - f(y_n)|$ does not tend to $0$ as $|x_n - y_n| \to 0$, $f(x) = x^2$ is not uniformly continuous on $[0, \infty)$.

Example 2: Topological Properties of Rational Numbers

Problem: Determine the interior, closure, boundary, and derived set of $\mathbb{Q}$ as a subset of $\mathbb{R}$ equipped with the standard topology.

Solution:

• Interior: Every non-empty open interval $(a, b) \subset \mathbb{R}$ contains irrational numbers, so no open interval can be contained in $\mathbb{Q}$. Hence, $\text{Int}(\mathbb{Q}) = \emptyset$.
• Closure: Since every real number $x \in \mathbb{R}$ is a limit point of rational numbers (density of $\mathbb{Q}$ in $\mathbb{R}$), every open ball around $x$ intersects $\mathbb{Q}$. Hence, $\overline{\mathbb{Q}} = \mathbb{R}$.
• Derived Set: The set of all limit points $\mathbb{Q}' = \mathbb{R}$.
• Boundary: $\partial \mathbb{Q} = \overline{\mathbb{Q}} \setminus \text{Int}(\mathbb{Q}) = \mathbb{R} \setminus \emptyset = \mathbb{R}$.

Example 3: Verifying the Hausdorff ($T_2$) Property

Problem: Prove that any metric space $(X, d)$ with its induced topology is a Hausdorff ($T_2$) space.

Solution:

Let $x, y \in X$ be distinct points such that $x \neq y$. Then $d(x, y) = r > 0$.
Define open balls $U = B\left(x, \frac{r}{2}\right)$ and $V = B\left(y, \frac{r}{2}\right)$.
Suppose there exists $z \in U \cap V$. By the triangle inequality: $$d(x, y) \le d(x, z) + d(z, y) < \frac{r}{2} + \frac{r}{2} = r$$ This leads to $r < r$, which is a contradiction. Hence, $U \cap V = \emptyset$. Since $x$ and $y$ have disjoint open neighborhoods, the space is $T_2$.

Example 4: Pointwise vs. Uniform Convergence of Functions

Problem: Analyze the convergence of $f_n(x) = x^n$ on $E = [0, 1]$.

Solution:

The pointwise limit function is: $$f(x) = \lim_{n \to \infty} x^n = \begin{cases} 0 & \text{if } 0 \le x < 1 \\ 1 & \text{if } x = 1 \end{cases}$$ Each $f_n(x)$ is continuous on $[0, 1]$, but the pointwise limit $f(x)$ is discontinuous at $x = 1$. By the Uniform Limit Theorem (which states that the uniform limit of continuous functions must be continuous), the sequence $(f_n)$ cannot converge uniformly on $[0, 1]$.

Part 3: High-Yield Solved MCQs (PPSC Exam Style)

1. Which of the following subsets of $\mathbb{R}$ is compact with respect to the usual topology?
   (A) $(0, 1]$
   (B) $\mathbb{Q} \cap [0, 1]$
   (C) $\{ \frac{1}{n} : n \in \mathbb{N} \} \cup \{0\}$
   (D) $(-\infty, 5]$
   
   Correct Answer: (C)
   Explanation: According to the Heine-Borel Theorem, a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded.
   • $(0, 1]$ is not closed because $0$ is a limit point not in the set.
   • $\mathbb{Q} \cap [0, 1]$ is not closed because its closure is $[0, 1]$.
   • $\{-\infty, 5]$ is bounded below by no real number (unbounded).
   • $A = \{\frac{1}{n} : n \in \mathbb{N}\} \cup \{0\}$ contains its only limit point $0$, so it is closed. It is also clearly bounded in $[0, 1]$. Thus, it is compact.


2. Let $X$ be an infinite set with the cofinite topology $\tau_{co}$. Then $X$ is:
   (A) Always $T_2$ (Hausdorff)
   (B) Always $T_1$ but never $T_2$
   (C) Neither $T_0$ nor $T_1$
   (D) $T_4$ (Normal)
   
   Correct Answer: (B)
   Explanation: In a cofinite topology on an infinite set $X$, every singleton set $\{x\}$ is closed because $X \setminus \{x\}$ has a finite complement, making it open. Therefore, the space is $T_1$. However, any two non-empty open sets $U, V$ must intersect because if $U \cap V = \emptyset$, then $X = U^c \cup V^c$, which implies $X$ is a union of two finite sets—a contradiction since $X$ is infinite. Hence, no two disjoint open sets exist, making it non-$T_2$.


3. The set of limit points (derived set) of $A = \{\frac{1}{n} + \frac{1}{m} : n, m \in \mathbb{N}\}$ in $\mathbb{R}$ is:
   (A) $\{0\}$
   (B) $\{\frac{1}{n} : n \in \mathbb{N}\ me\}$
   (C) $\{\frac{1}{n} : n \in \mathbb{N}\} \cup \{0\}$
   (D) $\emptyset$
   
   Correct Answer: (C)
   Explanation: Fixing $n$ and letting $m \to \infty$, the sequence converges to $\frac{1}{n}$. Thus, every point of the form $\frac{1}{n}$ is a limit point. Furthermore, letting both $n, m \to \infty$ yields $0$ as a limit point. Hence, the set of all limit points $A' = \{\frac{1}{n} : n \in \mathbb{N}\} \cup \{0\}$.


4. Let $f: X \to Y$ be a continuous bijective mapping from a compact space $X$ onto a Hausdorff space $Y$. Then $f$ is a:
   (A) Homeomorphism
   (B) Non-open mapping
   (C) Discontinuous mapping
   (D) Constant mapping
   
   Correct Answer: (A)
   Explanation: A fundamental theorem in topology states that if $f: X \to Y$ is continuous, $X$ is compact, and $Y$ is Hausdorff, then $f$ is a closed map. Since $f$ is a closed continuous bijection, its inverse $f^{-1}$ is continuous, which makes $f$ a homeomorphism.


5. Consider the series of functions $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2}$. The series converges:
   (A) Pointwise only on $[0, 2\pi]$
   (B) Uniformly on $\mathbb{R}$ by the Weierstrass M-Test
   (C) Divergent on $\mathbb{R}$
   (D) Uniformly on $(0, \pi)$ only
   
   Correct Answer: (B)
   Explanation: We have $|u_n(x)| = \left|\frac{\sin(nx)}{n^2}\right| \le \frac{1}{n^2} = M_n$ for all $x \in \mathbb{R}$. The numerical series $\sum_{n=1}^{\infty} M_n = \sum_{n=1}^{\infty} \frac{1}{n^2}$ is a $p$-series with $p = 2 > 1$, which converges. By the Weierstrass M-Test, the series converges uniformly and absolutely on all of $\mathbb{R}$.


6. Which of the following spaces is always connected?
   (A) $\mathbb{Q}$ with the induced topology from $\mathbb{R}$
   (B) Any discrete space with more than one point
   (C) The set $\mathbb{R}$ with the lower limit topology (Sorgenfrey line)
   (D) The interval $[a, b] \subset \mathbb{R}$ with the usual topology
   
   Correct Answer: (D)
   Explanation: In $\mathbb{R}$ equipped with the standard topology, a subset is connected if and only if it is an interval. Thus $[a, b]$ is connected. $\mathbb{Q}$ is totally disconnected, discrete spaces with $>1$ point are disconnected, and the Sorgenfrey line is totally disconnected.


7. If $A$ and $B$ are dense subsets of a topological space $X$, then:
   (A) $A \cap B$ is always dense in $X$
   (B) $A \cup B$ is dense in $X$
   (C) $A \setminus B$ is empty
   (D) $A \cap B$ is empty
   
   Correct Answer: (B)
   Explanation: Since $A \subset A \cup B$, taking closures yields $X = \overline{A} \subset \overline{A \cup B} \subset X$, so $\overline{A \cup B} = X$. Hence $A \cup B$ is dense. Notice that $A \cap B$ is not necessarily dense (e.g., $A = \mathbb{Q}$ and $B = \mathbb{R} \setminus \mathbb{Q}$ are both dense in $\mathbb{R}$, but $\mathbb{Q} \cap (\mathbb{R} \setminus \mathbb{Q}) = \emptyset$).


8. A topological space $X$ is regular if and only if for every point $x \in X$ and open set $U$ containing $x$, there exists an open set $V$ such that:
   (A) $x \in V \subset \overline{V} \subset U$
   (B) $x \in \overline{V} \subset V \subset U$
   (C) $V \cap U = \emptyset$
   (D) $\overline{V} = U$
   
   Correct Answer: (A)
   Explanation: This is a standard equivalent formulation of regularity ($T_3$ property). A space $X$ is regular if for every closed set $F$ and point $x \notin F$, there exist disjoint open sets separating them. Equivalently, every point $x$ has an open neighborhood $V$ whose closure $\overline{V}$ is entirely contained within an open neighborhood $U$ of $x$, i.e., $x \in V \subset \overline{V} \subset U$.

Summary Checklist for PPSC Aspirants

• Review Cantor's Intersection Theorem, Baire Category Theorem, and Cauchy Criterion for Real Analysis.
• Revise First/Second Countability, Separation Axioms hierarchy ($T_0 \Leftarrow T_1 \Leftarrow T_2 \Leftarrow T_3 \Leftarrow T_4$), and Connectedness theorems for Topology.
• Practice speed-solving past PPSC Lecturer papers (focus on 2011, 2015, 2017, 2020, and 2022 papers).