Real Analysis

Closed sets

A closed set in a metric space is one whose complement is open, or equivalently, a set that contains all its limit points. This topic covers closed sets and their topological characterizations.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

📖 Topic progress

○ Mark as complete

Definition and statement

Characterization of Closed Sets via Limit Points

A subset $F$ of a metric space $(X, d)$ is closed if and only if every limit point of $F$ belongs to $F$. That is, if $F'$ denotes the derived set of $F$, then $F' \subset F$.

📐 Geometric Interpretation

Geometrically, closed sets contain their boundaries and 'accumulation zones', ensuring that no sequence of points within the set can converge to a point outside the set.

⚙️ Algebraic & Structural Meaning

Algebraically, closed sets are stable under arbitrary intersections and finite unions, serving as the foundational building blocks for defining continuous mappings and compact spaces.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Metric Space Structure	`(X, d) Metric Space`	Strictly Mandatory / Necessary	Without a metric space or topological space, notions of limits, open balls, and convergence cannot be rigorously defined.
Complementary Openness	`X \setminus F \text{ is open}`	Strictly Mandatory	If the complement is not open, the set fails to contain its boundary points.
Sequential Closure	`Sequential limit condition`	Essential for characterization	Failing to account for all convergent sequences leads to missing boundary accumulation points.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Finite Unions and Intersections

$$\text{Finite union of closed sets is closed; Arbitrary intersection of closed sets is closed.}$

Rule: Never waste time checking limits for finite unions or arbitrary intersections; apply topological closure laws instantly.

Application: Directly tested in UP PGT/TGT for verifying statement correctness.

Singleton Sets in Metric Spaces

$$ \forall x \in X, \{x\} \text{ is closed in any metric space.} $$

Rule: Every singleton set is closed because its complement is open (or it has no limit points).

Application: Used to quickly deduce that finite sets are closed.

Continuous Image of Closed Sets

$$ f: X \to Y \text{ continuous}, \ F \subset Y \text{ closed} \implies f^{-1}(F) \text{ is closed in } X. $$

Rule: Use inverse images under continuous functions to test closedness without epsilon-delta proofs.

Closed Interval Property in $\mathbb{R}$

$$ [a, b], [a, \infty), (-\infty, b], \text{ and } \{a\} \text{ are closed in } \mathbb{R}. $$

Rule: Any interval containing its endpoints in $\mathbb{R}$ with the usual metric is closed.

Application: Fundamental baseline for real analysis MCQs.

Closure Operator Identity

$$ F \text{ is closed} \iff \bar{F} = F. $$

Rule: To prove or test if a set is closed, check if it equals its topological closure.

Rapid Elimination Trick for Open/Closed Hybrid Sets

$$ (a, b] \text{ is neither open nor closed in } \mathbb{R}. $$

Rule: Sets that are half-open/half-closed are immediately eliminated from purely closed or open options.

Application: Saves 30 seconds on classification questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Arbitrary Union of Closed Sets

Common Pitfall: Assuming that an infinite (arbitrary) union of closed sets is always closed.

Counterexample: $$\bigcup_{n=1}^{\infty} \left[0, 1 - \frac{1}{n}\right] = [0, 1)$$

Why it fails: The arbitrary union of closed intervals yields a half-open interval $[0, 1)$, which is NOT closed in $\mathbb{R}$.

🛡️ Exam Defense: Remember: Finite union of closed sets is closed, but infinite union can be open, closed, or neither.

⚠️ COMMON ERROR #02

Confusing Closure with Bounddness

Common Pitfall: Assuming bounded sets are closed or closed sets must be bounded.

Counterexample: $$[0, \infty) \text{ is closed in } \mathbb{R} \text{ but unbounded. } (0, 1] \text{ is bounded but not closed.}$

Why it fails: Closedness is a topological property related to limits/complements, whereas boundedness is a metric property.

🛡️ Exam Defense: Never conflate boundedness with closedness.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Closed Set	Complement is open	F^c \text{ is open}	Contains all boundary points	Includes endpoints like [a, b]
Open Set	Complement is closed	F \text{ is closed}	Contains an open ball around each point	Strict inequalities like (a, b)
Closure (\bar{F})	Subset of metric space	\bar{F} = F \cup F'	Set augmented with all its limit points	Smallest closed set containing F
Derived Set (F')	Subset of metric space	Set of all limit points	Points accumulation locus	Points approached by sequences in F

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Which of the following sets is closed in the usual metric space $\mathbb{R}$?

(A) \mathbb{Q}(B) (0, 1](C) \{x \in \mathbb{R} : x^2 \le 2\}(D) \{1/n : n \in \mathbb{N}\}

📚 Method A: Formal Textbook Derivation

Step 1: Note that $\{x \in \mathbb{R} : x^2 \le 2\} = [-\sqrt{2}, \sqrt{2}]$. Step 2: Since it is a closed interval, it contains all its limit points and is closed. Step 3: $\mathbb{Q}$ is neither open nor closed, $(0, 1]$ is half-open, and $\mathbb{N}^* = \{1/n\}$ lacks 0 which is a limit point.

⚡ Method B: 25-Second Exam Shortcut

Recognize the set as the inverse image of the closed set $[0, 2]$ under the continuous function $f(x) = x^2$.

💡 Tip: Look for inequalities involving $\le$ or $\ge$ with continuous functions.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $S = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}$. In the usual metric space $\mathbb{R}$, the set $S$

(A) is closed(B) is open(C) has 0 as a limit point but is not closed(D) is compact

📚 Method A: Formal Textbook Derivation

Step 1: The sequence $\{1/n\}$ converges to $0$. Step 2: Thus $0$ is a limit point of $S$. Step 3: However, $0 \notin S$. Step 4: Since $S'$ contains $0$ but $0 \notin S$, $S$ is not closed.

⚡ Method B: 25-Second Exam Shortcut

A countable set with a single limit point outside the set is a classic non-closed set.

💡 Tip: Check limit points of sequences of the form $1/n$.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following is always a closed set in any metric space $(X, d)$?

A The whole space $X$

B Any finite union of open sets

C Any countable intersection of open sets

D Any subset

Derivation & Analysis:

The whole space $X$ and the empty set $\emptyset$ are both open and closed (clopen) in any metric space.

MCQ 2: In the metric space $\mathbb{R}$ with usual metric, the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is:

A Closed

B Open

C Neither open nor closed

D Compact

Derivation & Analysis:

$\mathbb{R} \setminus \mathbb{Q}$ is neither open nor closed. Its interior is empty and its closure is $\mathbb{R}$.

MCQ 3: The intersection of an arbitrary family of closed sets in a metric space is:

A Always open

B Always closed

C Never closed

D Finite

Derivation & Analysis:

By De Morgan's laws, the complement of an arbitrary intersection of closed sets is an arbitrary union of open sets, which is open. Thus, the intersection is closed.

MCQ 4: Let $F_n = \left[0, 1 + \frac{1}{n}\right]$ for $n \in \mathbb{N}$. Then $\bigcap_{n=1}^{\infty} F_n$ is:

A (0, 1)

B [0, 1]

C [0, 1)

D (0, 1]

Derivation & Analysis:

The intersection of nested closed intervals $[0, 1 + 1/n]$ yields the closed interval $[0, 1]$.

MCQ 5: Which of the following sets is closed in $\mathbb{R}$?

A $(0, \infty)$

B $\mathbb{Z}$

C $\mathbb{Q}$

D $(-\infty, 0) \cup (0, \infty)$

Derivation & Analysis:

$\mathbb{Z}$ is closed in $\mathbb{R}$ because its complement $\mathbb{R} \setminus \mathbb{Z}$ is a union of open intervals $(n, n+1)$, which is open.

MCQ 6: If $A$ and $B$ are closed subsets of a metric space $X$, then $A \cup B$ is:

A Open

B Closed

C Neither open nor closed

D Empty

Derivation & Analysis:

A finite union of closed sets is always closed.

MCQ 7: The boundary $\partial S$ of any subset $S$ of a metric space is always:

A An open set

B A closed set

C A compact set

D A countable set

Derivation & Analysis:

The boundary of any set $S$, defined as $\bar{S} \setminus S^o$, is always a closed set because it is the intersection of two closed sets (closure and complement of interior).

MCQ 8: If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function, then the set $\{x \in \mathbb{R} : f(x) = 0\}$ is:

A Open

B Closed

C Bounded

D Compact

Derivation & Analysis:

The set $\{x : f(x) = 0\} = f^{-1}(\{0\})$. Since $\{0\}$ is closed in $\mathbb{R}$ and $f$ is continuous, its inverse image is closed.

MCQ 9: In the discrete metric space $(X, d)$, every subset is:

A Open only

B Closed only

C Both open and closed (clopen)

D Neither open nor closed

Derivation & Analysis:

In a discrete metric space, every singleton is open, hence every subset (being a union of singletons) is open, and its complement is also open, making every subset closed as well.

Q10

MCQ 10: Which of the following statements is FALSE?

A Every closed set is the complement of an open set.

B The closure of any set is closed.

C The union of infinitely many closed sets is always closed.

D Every singleton set in $\mathbb{R}$ is closed.

Derivation & Analysis:

The union of infinitely many closed sets need not be closed (e.g., $\bigcup [0, 1 - 1/n] = [0, 1)$).

Common questions

Yes! In any metric space $X$, both the empty set $\emptyset$ and the entire space $X$ are always both open and closed (often called 'clopen'). In disconnected spaces, other non-trivial clopen sets may also exist.

Topological axioms only guarantee that finite unions of closed sets are closed. Infinite (arbitrary) unions can expand the set to include boundary points that form an open or half-open boundary, as seen in $\bigcup_{n=1}^\infty [0, 1 - 1/n] = [0, 1)$. linkage.

Check if the inequality is non-strict (using $\le$ or $\ge$). Sets defined by continuous functions with non-strict inequalities (e.g., $f(x) \le c$) are closed, whereas strict inequalities ($f(x) < c$) define open sets.