Real Analysis

Derived set

Q: FAQ 1: Can a derived set be larger than the original set?

Yes, absolutely. For instance, if $A = \mathbb{Q}$ (the set of rational numbers) in $\mathbb{R}$, its derived set $A'$ is $\mathbb{R}$. Here, the derived set is strictly larger than the original set.

Q: FAQ 2: What is the most common reason students lose marks on this topic in UP PGT exams?

Candidates frequently confuse isolated points with limit points, or fail to recognize that limit points do not necessarily belong to the original set. Another frequent trap is miscalculating derived sets of multi-component sets or double-sequence sets.

Q: FAQ 3: How can candidates solve derived set questions faster during the exam?

Memorize standard benchmark results: finite sets have empty derived sets, intervals $(a, b)$ have closed intervals $[a, b]$ as derived sets, $\mathbb{Q}' = \mathbb{R}$, and $\mathbb{Z}' = \emptyset$. Utilizing these standard templates allows you to solve questions.

The derived set of a subset in a metric space is the collection of all its accumulation points (or limit points), serving as the foundational topological tool for analyzing closure, density, and compactness.

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

📖 Topic progress

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Definition and statement

Definition of Derived Set and Accumulation Point

Let $(X, d)$ be a metric space and $A \subset X$. A point $x \in X$ is called a limit point (or accumulation point) of $A$ if for every $\epsilon > 0$, the open ball $B_\epsilon(x)$ contains at least one point of $A$ distinct from $x$, i.e., $(B_\epsilon(x) \setminus \{x\}) \cap A \neq \emptyset$. The set of all limit points of $A$ is denoted by $A'$ and is called the derived set of $A$.

📐 Geometric Interpretation

Geometrically, $x \in A'$ if and only if every neighborhood around $x$ infinitely crowds points of $A$ (or at least contains points of $A$ other than $x$ itself). The point $x$ does not necessarily belong to $A$; it can sit on the boundary or in the exterior.

⚙️ Algebraic & Structural Meaning

Algebraically, $x \in A'$ if and only if there exists a sequence $(x_n)$ in $A$ such that $x_n \neq x$ for all $n$ and $\lim_{n \to \infty} x_n = x$. This sequence-theoretic characterization bridges topology with real analysis limit operations.

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 $(X, d)$	`General topological/metric space`	Strictly Mandatory / Necessary	Without a distance function or topology, the concept of a neighborhood $B_\epsilon(x)$ is undefined, making limit points meaningless.
Exclusion of Center Point in Neighborhood	`Neighborhood condition $B_\epsilon(x) \setminus \{x\}`	Strictly Mandatory	If the point $x$ itself were allowed to satisfy the intersection condition unconditionally, every isolated point of $A$ would erroneously become a limit point.
Non-emptiness of Subset $A$	`Subset $A \subset X$`	Essential for non-trivial derived set	If $A = \emptyset$, its derived set $A' = \emptyset$, as no points exist to form limit points.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Derived Set of Finite Sets

$$ \text{If } A \text{ is finite, then } A' = \emptyset $$

Rule: Any finite subset of $\mathbb{R}$ has an empty derived set because isolated points cannot have accumulation points.

Application: Frequently asked in UP PGT/TGT to test basic set topology definitions.

Derived Set of Standard Intervals

$$ (a, b)' = [a, b]', [a, b)' = [a, b], \text{ and } (a, b] = [a, b] $$

Rule: The derived set of any open, closed, or semi-open interval with endpoints $a < b$ is precisely the corresponding closed interval $[a, b]$.

Derived Set of $\mathbb{Q}$ and $\mathbb{Z}$

$$ \mathbb{Q}' = \mathbb{R}, \quad \mathbb{Z}' = \emptyset, \quad \mathbb{N}' = \emptyset $$

Rule: Dense sets in $\mathbb{R}$ have $\mathbb{R}$ as their derived set, whereas discrete sets have empty derived sets.

Application: Core conceptual question in DSSSB PGT and UP PGT.

Cantor Set Derived Set

$$ \mathcal{C}' = \mathcal{C} $$

Rule: The Cantor ternary set $\mathcal{C}$ is a perfect set, meaning it equals its own derived set ($\mathcal{C} = \mathcal{C}'$).

Union of Derived Sets

$$ (A \cup B)' = A' \cup B' $$

Rule: The derived set distributes perfectly across finite unions.

Infinite Countable Bounded Sets

$$ \text{If } A = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}, \text{ then } A' = \{0\} $$

Rule: For a sequence converging to a limit $L$ with distinct terms, the derived set consists solely of the limit point ${L}$.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Isolated Points with Limit Points

Common Pitfall: Candidates assume that points belonging to set $A$ must belong to its derived set $A'$.

Counterexample: Let $A = \{1, 2, 3\}$. Here $A = \{1, 2, 3\}$, but $A' = \emptyset$.

Why it fails: None of the points in $A$ have arbitrarily close neighboring points belonging to $A$. They are isolated points.

🛡️ Exam Defense: Remember: $A'$ contains limit points, which may or may not belong to $A$. An element of $A$ can be isolated.

⚠️ COMMON ERROR #02

Infinite Union vs Infinite Intersection Rule

Common Pitfall: Assuming infinite unions or infinite intersections behave nicely with derived sets, such as claiming $(\bigcap A_n)' = \bigcap A_n'$.

Counterexample: Let $A_n = (0, \frac{1}{n})$, whose intersection is empty, but infinite union properties can also fail for derived sets.

Why it fails: While $(A \cup B)' = A' \cup B'$ holds for finite unions, infinite unions satisfy only $\bigcup A_n' \subset (\bigcup A_n)'$, not equality in general.

🛡️ Exam Defense: Never distribute derived sets over infinite unions or intersections without rigorous verification.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Derived Set ($A'$)	Subset $A$ of metric space $X$	$A' = \{x \in X : B_\epsilon(x) \cap A \setminus \{x\} \neq \emptyset\}$	Set of all accumulation points	Look for limit points / clustering behavior
Closure ($\overline{A}$)	Subset $A$ of metric space $X$	$\overline{A} = A \cup A'$	Set union with all its limit points (closed hull)	Look for smallest closed set containing $A$
Isolated Point	Point $x \in A$	$\exists \, \epsilon > 0$ such that $B_\epsilon(x) \cap A = \{x\}$	Point separated from the rest of the set	Finite sets, integer points in $\mathbb{R}$
Perfect Set	Subset $A$ of metric space $X$	$A = A'$ ($A$ is closed and $A \subset A'$)	No isolated points; every point is an accumulation point	Cantor set, closed intervals $[a, b]$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the derived set of the set $S = \left\{ \frac{1}{m} + \frac{1}{n} : m, n \in \mathbb{N} \right\}$ in $\mathbb{R}$.

(A) $\left\{ \frac{1}{m} : m \in \mathbb{N} \right\} \cup \{0\}$(B) $\left\{ \frac{1}{m} : m \in \mathbb{N} \right\}$(C) $\left\{ 0 \right\}$(D) $\mathbb{R}$

📚 Method A: Formal Textbook Derivation

Step 1: Fix $m$ and let $n \to \infty$, giving limit points $\frac{1}{m}$ for each $m \in \mathbb{N}$. Step 2: Let both $m \to \infty$ and $n \to \infty$, giving the limit point $0$. Step 3: Combine these limits to form the complete derived set $S' = \left\{ \frac{1}{m} : m \in \mathbb{N} \right\} \cup \{0\}$.

⚡ Method B: 25-Second Exam Shortcut

Fix one variable to infinity and inspect boundary limits, then let both go to infinity simultaneously.

💡 Tip: Double-variable sequence sets always produce limits corresponding to fixing one variable plus the origin.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $A = \mathbb{Q} \cap [0, 1]$, what is the derived set $A'$?

(A) $\mathbb{Q} \cap [0, 1]$(B) $[0, 1]$(C) $(0, 1)$(D) $\emptyset$

📚 Method A: Formal Textbook Derivation

Step 1: Recognize that $\mathbb{Q}$ is dense in $\mathbb{R}$. Step 2: Every real number in $[0, 1]$ can be expressed as the limit of a sequence of distinct rational numbers in $[0, 1]$. Step 3: Therefore, every point of $[0, 1]$ is a limit point of $A$, giving $A' = [0, 1]$.

⚡ Method B: 25-Second Exam Shortcut

The derived set of any dense subset of an interval is the entire closed interval.

💡 Tip: Remember that the derived set of rational numbers in an interval is the closed interval, even though the set itself consists only of rationals.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the derived set of the set of integers $\mathbb{Z}$ in the standard metric space $\mathbb{R}$?

A $\mathbb{Z}$

B $\mathbb{R}$

C $\emptyset$

D $\mathbb{N}$

Derivation & Analysis:

Every integer $n \in \mathbb{Z}$ is an isolated point because the open interval $(n-1/2, n+1/2)$ contains no other integers. Thus, $\mathbb{Z}' = \emptyset$.

MCQ 2: If $A = (0, 1) \cup \{2\}$, what is the derived set $A'$?

A $(0, 1)$

B $[0, 1]$

C $[0, 1] \cup \{2\}$

D $[0, 1]$

Derivation & Analysis:

The derived set of $(0, 1)$ is $[0, 1]$. The singleton set $\{2\}$ has an empty derived set. By additivity of derived sets over finite unions, $A' = (0, 1)' \cup \{2\}' = [0, 1] \cup \emptyset = [0, 1].

MCQ 3: Which of the following sets is equal to its own derived set (i.e., a perfect set)?

A Finite set

B The set of rational numbers $\mathbb{Q}$

C Closed interval $[a, b]$

D Set of natural numbers $\mathbb{N}$

Derivation & Analysis:

A closed interval $[a, b]$ is closed and has no isolated points, so its derived set is $[a, b]$ itself. Thus, it is a perfect set.

MCQ 4: Let $A$ be a subset of a metric space $X$. Which of the following relations between closure $\overline{A}$ and derived set $A'$ is always true?

A $\overline{A} = A \cup A'$

B $\overline{A} = A'$

C $\overline{A} \cap A' = \emptyset$

D $\overline{A} \subset A'$

Derivation & Analysis:

By definition, the closure of a set $A$ is the union of the set itself and its set of limit points: $\overline{A} = A \cup A'$.

MCQ 5: What is the derived set of the set $A = \left\{ (-1)^n + \frac{1}{n} : n \in \mathbb{N} \right\}$?

A $\{-1, 1\}$

B $\{-1, 0, 1\}$

C $\{1\}$

D $\emptyset$

Derivation & Analysis:

The sequence has two limit points: as $n$ is even, terms approach $1$; as $n$ is odd, terms approach $-1$. Thus, the derived set is $\{-1, 1\}$.

MCQ 6: If $A$ and $B$ are two subsets of a metric space $X$, then which of the following is always true?

A $(A \cup B)' = A' \cap B'$

B $(A \cup B)' = A' \cup B'$

C $(A \cap B)' = A' \cap B'$

D $(A \cap B)' = A' \cup B'$

Derivation & Analysis:

The derived set operation distributes perfectly over finite unions: $(A \cup B)' = A' \cup B'$. For intersections, only $(A \cap B)' \subset A' \cap B'$ holds generally.

MCQ 7: What is the derived set of the set of all dyadic rationals $\mathcal{D} = \left\{ \frac{m}{2^n} : m \in \mathbb{Z}, n \in \nobreak \mathbb{N} \right\}$ in $\mathbb{R}$?

A $\mathcal{D}$

B $\mathbb{Q}$

C $\mathbb{R}$

D $\emptyset$

Derivation & Analysis:

Dyadic rationals are dense in $\mathbb{R}$, meaning every real number is a limit point of $\mathcal{D}$. Therefore, $\mathcal{D}' = \mathbb{R}$.

MCQ 8: Assertion (A): The derived set of every countable subset of $\mathbb{R}$ is countable. Reason (R): $\mathbb{Q}$ is a countable set, but its derived set is $\mathbb{R}$, which is uncountable.

A Both (A) and (R) are true and (R) is the correct explanation of (A).

B Both (A) and (R) are true but (R) is not the correct explanation of (A).

C (A) is true but (R) is false.

D (A) is false but (R) is true.

Derivation & Analysis:

Assertion (A) is false because $\mathbb{Q}$ is countable, yet its derived set is $\mathbb{R}$ (which is uncountable). Reason (R) correctly states this fact, so (A) is false and (R) is true.

MCQ 9: If $A'$ denotes the derived set of $A$, what is the derived set of the derived set, i.e., $(A')'$?

A It is always equal to $A'$.

B It is always a subset of $A'$.

C It is always a superset of $A'$.

D It is not necessarily related to $A'$.

Derivation & Analysis:

The derived set of a derived set is closed and satisfies $(A')' \subset A'$ for general subsets in metric spaces (known as Cantor-Bendixson property components).

Q10

MCQ 10: In the discrete metric space $(X, d)$ where $d(x, y) = 1$ if $x \neq y$ and $0$ if $x = y$, what is the derived set of any subset $A \subset X$?

A $A$

B $X$

C $\emptyset$

D The complement of $A$

Derivation & Analysis:

In a discrete metric space, every singleton set is an open ball of radius $< 1$. Thus, no point can have other points arbitrarily close to it, making $A' = \emptyset$ for every subset $A$.

Common questions

Yes, absolutely. For instance, if $A = \mathbb{Q}$ (the set of rational numbers) in $\mathbb{R}$, its derived set $A'$ is $\mathbb{R}$. Here, the derived set is strictly larger than the original set.

Candidates frequently confuse isolated points with limit points, or fail to recognize that limit points do not necessarily belong to the original set. Another frequent trap is miscalculating derived sets of multi-component sets or double-sequence sets.

Memorize standard benchmark results: finite sets have empty derived sets, intervals $(a, b)$ have closed intervals $[a, b]$ as derived sets, $\mathbb{Q}' = \mathbb{R}$, and $\mathbb{Z}' = \emptyset$. Utilizing these standard templates allows you to solve questions.