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Real Analysis

Open sets

An open set in a metric space is a subset whose every point is an interior point, admitting an open ball entirely contained within the set.

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Definition and statement

Definition and Characterization of Open Sets in $\mathbb{R}$
A subset $S \subset \mathbb{R}$ is said to be an open set if for every point $x \in S$, there exists a positive real number $\epsilon > 0$ such that the open interval $(x - \epsilon, x + \epsilon) \subset S$. Formally, $\forall x \in S, \, \exists \epsilon > 0$ such that $N_\epsilon(x) = (x - \epsilon, x + \epsilon) \subseteq S$.

📐 Geometric Interpretation

Geometrically, around every point in an open set, you can draw a small symmetric neighborhood (an open interval) that stays completely inside the set without poking outside its boundaries. This means open sets cannot contain their boundary points.

⚙️ Algebraic & Structural Meaning

Algebraically and topologically, arbitrary unions of open sets are open, and finite intersections of open sets are open. They allow the rigorous transition from discrete inequalities to continuous limits.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Metric / Distance Space Foundation Metric space $(X, d)$ Strictly Mandatory / Necessary Without a distance function or topology, the notion of 'nearness' and consequently 'interior points' is undefined.
Universal Quantification over Elements For all $x \in S$ Strictly Mandatory If even a single point in $S$ fails to have an $\epsilon$-neighborhood entirely contained in $S$, the set fails to be open.
Positivity of Neighborhood Radius Radius $\epsilon > 0$ Essential for non-degeneracy If $\epsilon = 0$, the neighborhood collapses to a single point, rendering interior verification impossible.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Open Intervals in $\mathbb{R}$

$$ (a, b) \text{ is open in } \mathbb{R}, \quad \forall a, b \in \mathbb{R} $$

Rule: Every open interval $(a, b)$, unbounded intervals $(a, \infty)$, $(-\infty, b)$, and the entire real line $\mathbb{R}$ are instantly recognized as open sets.

Application: Directly tested in UP PGT to identify topological properties of standard number subsets.
02

Arbitrary Union of Open Sets

$$ \bigcup_{\alpha \in I} U_\alpha \text{ is open if each } U_\alpha \text{ is open} $$

Rule: Any arbitrary (finite or infinite) union of open sets is always open. Look for union symbols in options for quick identification.

Application: Frequently framed in assertion-reason questions across DSSSB and KVS PGT.
03

Finite Intersection of Open Sets

$$ \bigcap_{i=1}^{n} U_i \text{ is open for finite } n $$

Rule: Finite intersection of open sets is open. Caution: Infinite intersection of open sets need not be open (e.g., $(-1/n, 1/n)$ intersection is $\{0\}$, which is closed).

Application: Classic trap question in UP TGT/PGT regarding countable intersections.
04

Complement of Closed Sets

$$ S \text{ is open} \iff S^c = \mathbb{R} \setminus S \text{ is closed} $$

Rule: To prove or test if a set is open, check if its complement is a standard closed set (like finite points or closed intervals).

Application: Used to eliminate lengthy interior point proofs in 15 seconds.
05

Discrete Metric Space Property

$$ \text{In }(X, d_{\text{discrete}}), \text{ every subset is open.} $$

Rule: Under the discrete metric, singleton sets and all subsets are both open and closed (clopen).

06

Rapid Elimination via Boundary Check

$$ S \text{ contains its boundary } \implies \text{ Not Open} $$

Rule: If a given set includes its endpoints or boundary points (such as $[a, b]$), instantly eliminate it from being an open set.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

The Infinite Intersection Fallacy

Common Pitfall: Assuming that infinite intersections of open sets are open, mirroring the rule for finite intersections.

Counterexample: $$ \bigcap_{n=1}^{\infty} \left(-\frac{1}{n}, \frac{1}{n}\right) = \{0\} $$

Why it fails: While each interval $(-1/n, 1/n)$ is open, their infinite intersection yields the singleton set $\{0\}$, which is closed in $\mathbb{R}$, not open.

🛡️ Exam Defense: Remember: Arbitrary union = Open; Finite intersection = Open; Infinite intersection = NOT necessarily open.
⚠️ COMMON ERROR #02

Confusing Closed Intervals with Open Sets

Common Pitfall: Thinking that a set like $[0, 1)$ is open because it lacks the right endpoint.

Counterexample: $0 \in [0, 1)$, but no open interval around $0$ lies entirely in $[0, 1)$

Why it fails: Any $\epsilon$-neighborhood around $0$, i.e., $(-\epsilon, \epsilon)$, contains negative numbers which fall outside $[0, 1)$.

🛡️ Exam Defense: Check both left and right neighborhoods at every boundary point. If the boundary is included, the set is not open.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Open SetEvery point is an interior point$S = \text{int}(S)$Surrounded entirely by local open neighborhoodsStrict inequalities like $(a, b)$ or $\mathbb{R} \setminus \text{closed}$
Closed SetContains all its limit points$S = \bar{S}$ or $S^c$ is openIncludes its boundaries and limitsNon-strict inequalities like $[a, b]$ or finite sets
Clopen SetBoth open and closed$S \text{ is open and } S \text{ is closed}$Component of disconnected spaceWhole space $\mathbb{R}$ or empty set $\emptyset$
Interior PointExists an open ball inside set$x \in N_\epsilon(x) \subset S$Deep inside the set, away from bordersPoints strictly satisfying set conditions

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Which of the following subsets of $\mathbb{R}$ is an open set?

(A) $\mathbb{Z}$ (Set of integers)(B) $(0, 1]$(C) $(-\infty, 5)$(D) $\{x \in \mathbb{R} : x \ge 0 \}$
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Check interval types. $\mathbb{Z}$ consists of isolated points with no interior intervals. $(0, 1]$ includes endpoint $1$. ${x \ge 0}$ includes $0$. Step 2: $(-\infty, 5)$ is a standard unbounded open interval. Step 3: Conclusion: $(-\infty, 5)$ is open.
⚡ Method B: 25-Second Exam Shortcut
Instantly recognize standard open intervals $(a, b)$ and $(-\infty, b)$. Eliminate sets with square brackets or discrete points.
💡 Tip: Look out for half-open intervals disguised in set-builder notation.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $S = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}$. In the standard topology of $\mathbb{R}$, the set $S$ is:

(A) An open set(B) A closed set(C) Neither open nor closed(D) Both open and closed
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Examine limit points of $S$. The limit point is $0$, but $0 \notin S$. Thus $S$ is not closed. Step 2: For any element $1/n \in S$, no open interval around $1/n$ is contained in $S$ because it contains irrational and rational numbers outside $S$. Thus $S$ is not open. Step 3: Conclusion: Neither open nor closed.
⚡ Method B: 25-Second Exam Shortcut
Countable sets of isolated points with a limit point outside the set are classic examples of neither open nor closed sets.
💡 Tip: Do not confuse the openness of $S$ with the openness of intervals. Singleton points in $\mathbb{R}$ are closed.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: Which of the following is always an open set in $\mathbb{R}$ with standard topology?
A The set of rational numbers $\mathbb{Q}$
B Any finite subset of $\mathbb{R}$
C An open interval $(a, b)$
D The closed interval $[a, b]$
Derivation & Analysis:

By definition, an open interval $(a, b)$ contains an open neighborhood around each of its points, making it an open set.

Q2
MCQ 2: The union of an arbitrary family of open sets is:
A Always a closed set
B Always an open set
C Sometimes finite
D Never open
Derivation & Analysis:

A foundational theorem in topology states that arbitrary union of open sets is always open.

Q3
MCQ 3: The intersection of infinitely many open sets in $\mathbb{R}$ is:
A Always open
B Always closed
C Always empty
D Not necessarily open
Derivation & Analysis:

Infinite intersections of open sets can yield closed sets, such as $\bigcap_{n=1}^\infty (-1/n, 1/n) = \{0\}$, which is closed.

Q4
MCQ 4: Which of the following sets is open in $\mathbb{R}$?
A $\mathbb{R} \setminus \{0\}$
B $\{0, 1, 2\}$
C $[0, 1)$
D $\mathbb{Q}$
Derivation & Analysis:

$\mathbb{R} \setminus \{0\}$ is the complement of a closed set (a singleton), hence it is open.

Q5
MCQ 5: The interior of the set $S = [1, 2]$ in $\mathbb{R}$ is:
A $[1, 2]$
B (1, 2)
C $[1, 2)$
D $\emptyset$
Derivation & Analysis:

The largest open set contained in $[1, 2]$ is the open interval $(1, 2)$.

Q6
MCQ 6: In a 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 space, singleton sets are open balls of radius $1$, making all subsets open and consequently closed.

Q7
MCQ 7: If $A$ and $B$ are open sets in $\mathbb{R}$, then $A \cap B$ is:
A Always open
B Always closed
C Empty
D Finite
Derivation & Analysis:

The intersection of a finite number of open sets (here two) is always open.

Q8
MCQ 8: Assertion (A): Every open interval is an open set. Reason (R): Every point in an open interval is an interior point.
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:

Reason directly defines why an open interval is an open set, satisfying the interior point condition for all its elements.

Q9
MCQ 9: The set of all irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ in $\mathbb{R}$ is:
A An open set
B A closed set
C Neither open nor closed
D Compact
Derivation & Analysis:

Neither $\mathbb{R} \setminus \mathbb{Q}$ nor its complement $\mathbb{Q}$ is open/closed alone; $\mathbb{Q}$ is dense and has empty interior, so its complement is dense and has empty interior.

Q10
MCQ 10: Which of the following is an open subset of $\mathbb{R}^2$?
A The unit circle $\{(x,y) : x^2 + y^2 = 1\}$
B The closed disc $\{(x,y) : x^2 + y^2 \le 1\}$
C The open disc $\{(x,y) : x^2 + y^2 < 1\}$
D The line segment $\{(x,0) : 0 \le x \le 1\}$
Derivation & Analysis:

An open disc defined by strict inequality ($< 1$) corresponds to an open ball in $\mathbb{R}^2$, hence it is an open set.

Common questions

Yes, the empty set is vacuously open because the condition 'for every element in $\emptyset$' holds true (there are no counterexamples). By duality, its complement $\mathbb{R}$ is closed, making $\emptyset$ both open and closed.

Students often confuse open intervals with open sets in general, failing to recognize that arbitrary infinite intersections of open sets can result in closed sets or sets that are neither open nor closed.

Use the complement rule: check if the complement of the given set is a standard closed set. Also, instantly eliminate any set defined with non-strict inequalities ($\le, \ge$) as candidates for open sets.

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