Analysis & Linear Algebra

Metric Spaces

6 previous-year questions →every option and the correct answer, free Practice this topic →one question at a time, no clock, check as you go

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1. Open/closed sets, limit points, closure, interior

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Lec-17 Open Sets

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Lec-18 Closure Points, Limit Points and isolated Points

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Distinguishes limit points from isolated points — a common exam confusion.

Lec-19 Closed sets

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Open and closed sets, closure, interior, boundary free

In plain words

Open and closed are not opposites. A set can be both, or neither. Any union of open sets is open, but only finite intersections are, and that word finite carries all the weight.

The operations

Arbitrary unions and finite intersections of open sets are open. ∩(−1/n, 1/n) = {0} shows infinite intersections fail.
Arbitrary intersections and finite unions of closed sets are closed. ∪[1/n, 1] = (0,1] shows infinite unions fail.
∅ and X are both open and closed (clopen). In a connected space these are the only ones.

Closure, interior, boundary

cl A = A ∪ A′ (A′ = limit points) = smallest closed set containing A.
int A = largest open set inside $A. \partial{}A =$ cl A ∖ int A.
A closed ⇔ A = cl A ⇔ A contains all its limit points ⇔ A contains all limits of its convergent sequences (in metric spaces).
A dense ⇔ cl A = X ⇔ every non-empty open set meets A.

Set in $\mathbb{R}$	int	cl	$\partial$
$\mathbb{Q}$	∅	$\mathbb{R}$	$\mathbb{R}$
(0,1)	(0,1)	[0,1]	{0,1}
$\mathbb{Z}$	∅	$\mathbb{Z}$	$\mathbb{Z}$
{1/n}	∅	{0} ∪ {1/n}	itself ∪ {0}
Cantor set	∅	itself	itself

Distributive failures

int(A ∪ B) ⊋ int A ∪ int B possible: $A = \mathbb{Q}, B = \mathbb{R}$ ∖ $\mathbb{Q}$ gives $\mathbb{R}$ vs ∅.
cl(A ∩ B) ⊊ cl A ∩ cl B possible: same example gives ∅ vs $\mathbb{R}$ .
The finite-index versions int(A ∩ B) = int A ∩ int B and cl(A ∪ B) = cl A ∪ cl B do hold.

Types of points

isolated (a neighbourhood meeting A only in itself), limit point, boundary point. Perfect = closed with no isolated points $(\mathbb{R}$ , Cantor set). A non-empty perfect subset of $\mathbb{R}$ is uncountable.

Key takeaways

Finite intersections keep open; finite unions keep closed.
Closure distributes over finite unions, interior over finite intersections — never the reverse. $- \mathbb{Q}$ vs its complement is the counterexample generator for all of these.

The unit ball, from diamond to squareinteractivefree

Exam focus: Sets can be both open and closed, or neither. Know which operations preserve openness (arbitrary unions, finite intersections) and the standard ℚ examples.

Practice

Part C2025 Juneshow ▾

Consider

\mathbb{R}

with the usual topology and S = {

a + b\sqrt{2}

a, b \in \mathbb{Q}

}

\subset \mathbb{R}

with the subspace topology. Which of the following statements are true?

A.S is dense in $\mathbb{R}$ .✓ Correct
B.S \ $\mathbb{Q}$ is dense in $\mathbb{R}$ .✓ Correct
C.S \ $\mathbb{Q}$ is discrete with subspace topology on S.
D.S is connected.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Let (X,d) be a metric space. For a non-empty subset A of X, and x∈X, define d(x,A) =

inf_{a\in{}A}

d(x,a). Which of the following statements are necessarily true?

A.For all x,y∈X and every non-empty subset A of X, we have d(x,A)−d(y,A) ≤ d(x,y).✓ Correct
B.For every non-empty subset A of X, the function x↦d(x,A) is uniformly continuous on X.✓ Correct
C.A non-empty subset A of X is closed if and only if d(x,A)>0 for all x in X outside A.✓ Correct
D.If X is compact and $C_{1},\dots,C_{k}$ are non-empty closed sets of X such that $\cap{}C_{i}=\emptyset$ , then the minimum value of $x\mapsto\sum{}d(x,C_{i})$ on X is 0.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider

\mathbb{R}^{2}

with the Euclidean metric d. Let

O = (0, 0), P_{0} = (0, 1) \in \mathbb{R}^{2}

, and for any integer

n \ge 1, P_{n} = (1/n, 1) \in \mathbb{R}^{2}

. Let X = ⋃

(n\ge0) L_{n}

, where

L_{n}

is the line segment joining

P_{n}

and O. Define

d_X : X \times X \to \mathbb{R}

as follows: d_X(a, b) = d(a, b) when

a, b \in L_{n}

for some n, and d_X(a, b) = d(a, O) + d(b, O) otherwise. Let

\tau

be the smallest topology such that the sets

B_(a,\varepsilon) =

{

b \in X : d_X(a, b) < \varepsilon

} are open in

\tau

for all a ∈ X and

\varepsilon > 0

. Which of the following statements are true?

A.d_X is a metric on X which induces the topology $\tau$ .✓ Correct
B.The topological space $(X, \tau)$ is connected.✓ Correct
C. $(P_{n})_{n}\ge_{1}$ converges to $P_{0}$ in the topological space $(X, \tau)$ .
D.The topological space $(X, \tau)$ is compact.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let

\mathbb{N}

be the set of positive integers. Consider

\mathbb{R}^{2}

with the Euclidean topology and the subsets A = {

(n, 1/n) : n \in \mathbb{N}

} and B = {

(n, 1/m) : n, m \in \mathbb{N}

}. Which of the following statements is true?

A.A is a closed subset of $\mathbb{R}^{2}$ but B is not a closed subset of $\mathbb{R}^{2}$ .✓ Correct
B.B is a closed subset of $\mathbb{R}^{2}$ but A is not a closed subset of $\mathbb{R}^{2}$ .
C.Both A and B are closed subsets of $\mathbb{R}^{2}$ .
D.Neither A nor B is a closed subset of $\mathbb{R}^{2}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

A, B \subseteq \mathbb{R}

. Which of the following always hold?

A.int(A ∩ B) = int(A) ∩ int(B)✓ Correct
B.int(A ∪ B) = int(A) ∪ int(B)
C.cl(A ∪ B) = cl(A) ∪ cl(B)✓ Correct
D. $\partial{}A$ is closed✓ Correct

Solution

(b) fails for $A = \mathbb{Q}, B = \mathbb{R}$ ∖ $\mathbb{Q}. \partial{}A =$ cl A ∩ cl(Aᶜ) is an intersection of closed sets.

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Part Cshow ▾

Let (X, d) be a metric space, A ⊆ X, and let A′ denote the set of limit points of A. Which of the following are always true?

A.A′ is closed.✓ Correct
B.The closure of A equals A ∪ A′.✓ Correct
C.If A is finite, then A′ = ∅.✓ Correct
D.A′ ⊆ A.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

\mathbb{R}

with the usual topology, the interior of

\mathbb{Q}

A. $\mathbb{Q}$ itself
B.∅✓ Correct
C. $\mathbb{R}$
D.the set of integers

The worked solution and the per-option analysis are in Trap Analysis.

2. Compactness: open covers, sequential, Heine–Borel

Read this as a chapter →

Lec-28 Compactness

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Open-cover definition first; sequential compactness comes later.

Lec-30 Characterizations of Compact Sets

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

The equivalences in metric spaces — exactly what Part C tests.

Lec-13 Weierstrass Theorem, Heine Borel Theorem, Connected set

NPTEL · A Basic Course in Real Analysis (IIT Kharagpur)

Heine–Borel, and why it is ℝⁿ-specific.

Compactness — the implication map free

In plain words

Compactness makes an infinite thing behave like a finite one, so a continuous function on it actually reaches its maximum. Closed and bounded is enough only inside $\mathbb{R}^n$ .

In a metric space the following are equivalent: (1) every open cover has a finite subcover, (2) every sequence has a convergent subsequence, (3) complete and totally bounded.

Exam rule: In $\mathbb{R}^{n}$ , compact ⇔ closed and bounded (Heine–Borel). Outside $\mathbb{R}^{n}$ , closed and bounded is not enough.

Spoiler	Closed?	Bounded?	Compact?
Closed unit ball in $\ell^{2}$	✓	✓	✗ $(e_{1}, e_{2}, \dots$ has no convergent subsequence)
$\mathbb{R}$ with the discrete metric	✓	✓	✗ (singletons form an open cover)
$[0,1] \cap \mathbb{Q}$	✗ in $\mathbb{R}$	✓	✗

compact ──▶ complete ──▶ closed (in any ambient space)
   └──▶ totally bounded ──▶ bounded
complete + totally bounded ⇔ compact   (always, in metric spaces)

Key takeaways

Three equivalent definitions; switch freely between them.
"Every continuous $f : X \to \mathbb{R}$ is bounded" ⇔ X compact (metric spaces).
Completeness is not topological; compactness is.

Why (0,1) is not compact — an explicit cover with no finite subcoverinteractivefree

Step 1 / 4The claim

$(0,1)$ is bounded, so if compactness were only about boundedness it would qualify. Show directly that it is not compact by exhibiting an open cover with no finite subcover.

Exam focus: Closed + bounded ⇒ compact ONLY in ℝⁿ. Know the ℓ² unit ball and discrete ℝ as spoilers. Compact ⇒ complete ⇒ closed.

Practice

Part C2025 Juneshow ▾

Let

p : \mathbb{R} \to \mathbb{R}

be a nonconstant polynomial. Which of the following statements are true?

A.The preimage of a compact set under p is a compact set.✓ Correct
B.The preimage of a connected set under p is a connected set.
C.Every point $x \in \mathbb{R}$ has an open neighbourhood U_x such that the restriction $p|_{U_x}$ is a homeomorphism onto an open set in $\mathbb{R}$ .
D.The image of a bounded set under p is a bounded set.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let (X, d) be a metric space. Which of the following are true?

A.X is compact iff every open cover of X has a finite subcover.✓ Correct
B.X is compact iff X is sequentially compact (every sequence has a convergent subsequence).✓ Correct
C.A compact metric space is complete and totally bounded.✓ Correct
D.Every closed and bounded subset of a metric space is compact.

Solution

(1) is the definition. (2) compactness and sequential compactness coincide in metric spaces. (3) is the standard characterisation. (4) is false in general: Heine–Borel (closed + bounded ⇒ compact) is special to $\mathbb{R}^{n}$ . In an infinite-dimensional space it fails — the closed unit ball of C[0,1] is closed and bounded but not compact.

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Part Bshow ▾

Which of the following subsets of

\mathbb{R}^{2}

is compact?

A.{ $(x, y) : x^{2} + y^{2} < 1$ }
B.{ $(x, y) : x^{2} + y^{2} \le 1$ }✓ Correct
C.{(x, y) : x ≥ 0, y ≥ 0}
D.{(x, y) : xy = 1}

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following subsets of

\mathbb{R}

, with the usual topology, are compact?

A.{0} ∪ { $1/n : n \in \mathbb{N}$ }✓ Correct
B. $\mathbb{Q} \cap [0, 1]$
C.The Cantor middle-third set✓ Correct
D.{ $1/n : n \in \mathbb{N}$ }

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let X = {

1/n : n \in \mathbb{N}

} ∪ {0} with the usual metric from

\mathbb{R}

. Which of the following are true?

A.X is compact✓ Correct
B.X is complete✓ Correct
C.Every function $f : X \to \mathbb{R}$ is continuous
D.X is connected

Solution

(a) True — closed and bounded in $\mathbb{R}$ .

(b) True — compact ⇒ complete.

(d) False — {1} is both open and closed in X.

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Part Cshow ▾

Let X = C[0, 1] with the sup-norm metric, and let B = {f ∈ X : ‖f‖ ≤ 1}. Which of the following are true?

A.B is closed and bounded in X.✓ Correct
B.B is not compact.✓ Correct
C.In any finite-dimensional normed space, the closed unit ball is compact.✓ Correct
D.The sequence $f_{n}(x) = x^{n}$ has a uniformly convergent subsequence in B.

Solution

(1) is immediate from the definition of B. (3) is Heine–Borel in finite dimensions. (2) B fails to be compact precisely because dimension is infinite — witnessed by (4) being false: $f_{n}(x) = x^{n}$ converges pointwise to a function that is 0 on [0,1) and 1 at x = 1, which is discontinuous, so no subsequence of continuous functions can converge to it uniformly.

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3. Completeness and Baire category

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Lec-21 Completeness

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Complete metric spaces and the standard complete/incomplete examples.

Lec-22 Baire Category Theorem

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Baire and its consequences — the source of several Part-C items.

Completeness, Banach fixed point and Baire category free

In plain words

Complete means no gaps: any sequence whose terms crowd together has something to crowd around. It depends on the distance you chose rather than the shape, so stretching a space can destroy it.

Completeness

Space	Complete?
$\mathbb{R}^{n}, \mathbb{C}^{n}, \ell$ ᵖ $(1 \le p \le \infty), C[a,b]$ with sup norm	✓
closed subset of a complete space	✓
$\mathbb{Q}, (0,1), \mathbb{R}$ ∖{0} (usual metric)	✗
C[0,1] with $\int$ \|f\| norm	✗ (step-function limits)
polynomials with sup norm on [0,1]	✗ (Weierstrass: closure is all of C[0,1])
any set with the discrete metric	✓ (Cauchy ⇒ eventually constant)
$(0, \infty)$ with d(x,y) = \|ln x − ln y\|	✓ (isometric to $\mathbb{R})$

Complete ⊆ X ⇒ closed in X. Closed ⊆ complete ⇒ complete.
Completeness is a metric, not topological, property $((0,1) \cong \mathbb{R})$ .
Every metric space has a completion, unique up to isometry.
Cantor intersection: complete ⇔ every decreasing sequence of non-empty closed sets with diameters → 0 has non-empty (singleton) intersection.

Banach contraction principle

T : X → X with d(Tx, Ty) ≤ k·d(x,y), k < 1, on a complete X ⇒ unique fixed point, and iterates converge from any start. Spoilers: T(x) = x + 1/x on $[1,\infty)$ satisfies d(Tx,Ty) < d(x,y) (not a strict contraction constant) and has no fixed point; x ↦ x/2 on (0,1] (incomplete) has none.

Baire category theorem

A complete metric space is not a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Consequences that appear as Part-C items: $- \mathbb{R}$ is uncountable; more generally a complete metric space with no isolated points is uncountable. $- \mathbb{Q}$ is not $a G_\delta$ in $\mathbb{R}$ ; the irrationals are not an $F_\sigma$ .

There is no function continuous exactly on $\mathbb{Q} ($ continuity set is always $G_\delta)$ .
A complete normed space cannot have a countably infinite Hamel basis (so $\mathbb{R}[x]$ with any norm is incomplete).
Pointwise limit of continuous functions is continuous on a dense $G_\delta$ .

Key takeaways

"Closed and inside something complete" is the fastest completeness proof; "has a Cauchy sequence escaping" is the fastest disproof.
Contraction needs k < 1 strictly and a complete space.
Baire turns "countable union of small sets" into an impossibility proof.

Baire category, and the argument it powersinteractive

This visual is part of Trap Analysis

Why ℝ is not a countable union of nowhere dense sets, and how that one fact proves several others.

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Exam focus: Completeness: closed subsets of complete spaces, ℓᵖ, C[0,1] with sup norm are complete; ℚ, (0,1), C[0,1] with L¹ norm are not. Baire: ℝ is not a countable union of nowhere dense sets; a complete metric space without isolated points is uncountable.

Practice

Part C2025 Juneshow ▾

Let C[0, 1] be the

\mathbb{R}-

vector space of real valued continuous functions equipped with the norm ‖f‖ =

sup_{x\in[0,1]}

|f(x)|. Let T : C[0, 1] → C[0, 1] be defined as

T(f)(x) = \int_{0}

ˣ f(t)dt, for x ∈ [0, 1]. Let

T^{n} = T

∘ T ∘ ⋯ ∘ T (n times). Which of the following statements are true?

A.There exists $\alpha \in (0, 1)$ such that for all f, g ∈ C[0, 1], ‖T(f) − T(g)‖ $\le \alpha$ ‖f − g‖.
B.There exists $\alpha \in (0, 1)$ such that for all f, g ∈ C[0, 1], ‖ $T^{2}(f) - T^{2}(g)$ ‖ $\le \alpha$ ‖f − g‖.✓ Correct
C.The set {f ∈ C[0, 1] : T(f) = f} is a singleton set.✓ Correct
D.‖ $T^{n}$ ‖ $\to \infty$ as $n \to \infty$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following statements about the Baire category theorem in

\mathbb{R}

are true?

A. $\mathbb{R}$ cannot be written as a countable union of nowhere dense sets.✓ Correct
B. $\mathbb{Q}$ is a countable union of nowhere dense subsets of $\mathbb{R}$ .✓ Correct
C.The irrationals form a set of first category in $\mathbb{R}$ .
D.A non-empty complete metric space with no isolated points is uncountable.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following follow from the Baire category theorem?

A. $\mathbb{R}$ ∖ $\mathbb{Q}$ is not a countable union of closed sets✓ Correct
B.There is no $f : \mathbb{R} \to \mathbb{R}$ continuous exactly at the rationals✓ Correct
C.Every infinite-dimensional Banach space has uncountable Hamel dimension✓ Correct
D. $\mathbb{Q}$ is a countable union of closed sets

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following metric spaces are complete?

A. $\mathbb{Q}$ with the usual metric
B.(0, 1] with the usual metric
C. $\mathbb{N}$ with the usual metric✓ Correct
D.C[0,1] with the metric $d(f,g) = \int_{0}^{1}$ |f − g|

Solution

$\mathbb{N}$ is closed in $\mathbb{R} ($ and discrete $). \mathbb{Q}$ and (0,1] have escaping Cauchy sequences; the $L^{1}$ metric on continuous functions is incomplete.

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Part Bshow ▾

Let

T : [1, \infty) \to [1, \infty), T(x) = x + 1/x

. Then

A.T is a contraction and has a unique fixed point
B.|T(x) − T(y)| < |x − y| for x ≠ y, but T has no fixed point✓ Correct
C.T has infinitely many fixed points
D.T is not continuous

The worked solution and the per-option analysis are in Trap Analysis.

4. Connectedness and path-connectedness

Read this as a chapter →

Lec-26 Connectedness

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Lec-27 Connected Sets

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Connected subsets of ℝ are exactly the intervals.

Connectedness — tests, preservation and the sine curve free

In plain words

Connected means you cannot cut it into two open pieces. Path-connected means you can walk between any two points. Walking is stronger, and one famous curve is connected without being walkable.

Definitions

X is connected if it is not a union of two disjoint non-empty open sets (equivalently, the only clopen sets are ∅ and X; equivalently every continuous f : X → {0,1} is constant). Path-connected: any two points joined by a continuous path.

Quick facts

Connected subsets of $\mathbb{R} =$ intervals (including points). Hence IVT.
Continuous image of a connected set is connected; of a path-connected set is path-connected.
Closure of a connected set is connected (any set between A and Ā is connected). Closure of a path-connected set need not be path-connected.
Union of connected sets with a common point is connected.
Product of connected spaces is connected.
Components are closed; path components need not be.
Path-connected ⇒ connected. Converse fails in general, holds for open subsets of $\mathbb{R}^{n}$ (and for locally path-connected spaces).

Standard examples

Space	Connected	Path-connected
$\mathbb{R}$ ∖{0}	✗	✗
$\mathbb{R}^{2}$ ∖{0}, $\mathbb{R}^{2}$ ∖(countable set)	✓	✓
$\mathbb{R}$ ∖ $\mathbb{Q}, \mathbb{Q}$	✗ (totally disconnected)	✗
GL $_n(\mathbb{R})$	✗ (two components by sign of det)	—
GL $_n(\mathbb{C})$ , SL $_n(\mathbb{R})$ , SO(n), U(n)	✓	✓
O(n)	✗ (two components)	—
{(x, sin 1/x): 0 < x ≤ 1} ∪ ({0}×[−1,1])	✓	✗
{(x,y): xy = 1}	✗	✗
Cantor set	✗ (totally disconnected)	✗
cofinite topology on an infinite set	✓	✓ (surprisingly)
Sorgenfrey line $\mathbb{R}_\ell$	✗ (totally disconnected)	✗

$\mathbb{R}$ vs $\mathbb{R}^{n}$

$\mathbb{R}$ and $\mathbb{R}^{n} (n \ge 2)$ are not homeomorphic: removing a point disconnects $\mathbb{R}$ but not $\mathbb{R}^{n}$ . Same trick: [0,1] vs the circle, [0,1) vs (0,1).

Key takeaways

Find a clopen set, or a continuous map onto {0,1}, to disprove connectedness.
"Connected but not path-connected" = topologist's sine curve; "open in $\mathbb{R}^{n}$ " removes the distinction.
Removing points is the standard homeomorphism obstruction.

Connected but not path-connectedinteractive

This visual is part of Trap Analysis

The topologist's sine curve, and why closure preserves connectedness but not path-connectedness.

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Exam focus: Connected subsets of ℝ are intervals; continuous images stay connected; path-connected ⇒ connected with the topologist's sine curve as the standard converse failure — except open subsets of ℝⁿ, where the two agree.

Practice

Part Cshow ▾

Which of the following subsets of

\mathbb{R}^{2} (

usual topology) are connected?

A.{(x, y) : xy = 1}
B.{ $(x, y) : x^{2} + y^{2} = 1$ }✓ Correct
C.{(x, y) : xy = 0}✓ Correct
D. $\mathbb{R}^{2}$ \ {(0, 0)}✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let (X, d) be a metric space. Which of the following are true?

A.If X is path-connected, then X is connected.✓ Correct
B.A convex subset of $\mathbb{R}^{n}$ , with the usual metric, is path-connected.✓ Correct
C.The union of two path-connected sets with a common point is path-connected.✓ Correct
D.Every connected subset of $\mathbb{R}^{n} (n \ge 2)$ is path-connected.

Solution

(1)–(3) are standard facts — (2) via straight-line paths, (3) by concatenating paths through the shared point.

(4) is false: it drops the hypothesis that makes the true theorem (every open connected subset of $\mathbb{R}^{n}$ is path-connected) work.

The topologist's sine curve is a connected subset of $\mathbb{R}^{2}$ that is not path-connected — it just isn't open.

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Part Cshow ▾

Which of the following are true?

A.Every open connected subset of $\mathbb{R}^{n}$ is path-connected✓ Correct
B.The closure of a path-connected set is path-connected
C.The continuous image of a connected set is connected✓ Correct
D.A space with the cofinite topology on an infinite set is connected✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let X = {(x, sin(1/x)) : 0 < x ≤ 1} ∪ ({0}

\times [-1, 1]) \subseteq \mathbb{R}^{2}

, the topologist's sine curve, with the Euclidean metric. Which of the following are true?

A.X is connected.✓ Correct
B.X is compact.✓ Correct
C.X is path-connected.
D.Every connected component of X is a single point.

Solution

(1) X is the closure of the connected curve piece, and the closure of a connected set is connected.

(2) X is closed (it already equals its own closure) and bounded, hence compact by Heine–Borel.

(3) is false: no path from the oscillating curve can reach the segment {0} × [−1, 1] — the standard failure of path-connectedness.

(4) is false directly from (1): X is itself connected, so it has exactly one component, not many single-point ones.

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Part Bshow ▾

Let X = {(x, sin(1/x)) : 0 < x ≤ 1} ∪ ({0}

\times [-1, 1]) \subset \mathbb{R}^{2}

. Then X is

A.connected and path-connected
B.connected but not path-connected✓ Correct
C.path-connected but not connected
D.neither connected nor path-connected

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following are connected?

A.GL $_{2}(\mathbb{R})$
B.SL $_{2}(\mathbb{R})$ ✓ Correct
C. $\mathbb{R}^{2}$ ∖ $\mathbb{Q}^{2}$ ✓ Correct
D.O(2), the group of 2×2 real orthogonal matrices

Solution

(a) det < 0 and det > 0 are disjoint open pieces.

(b) SL $_{2}(\mathbb{R})$ is path-connected.

(c) the complement of a countable set in $\mathbb{R}^{2}$ is path-connected: through any two points pass uncountably many lines, avoid the countably many bad ones.

(d) O(2) splits by det = ±1.

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Metric Spaces

1. Open/closed sets, limit points, closure, interior

Open and closed sets, closure, interior, boundary free

In plain words

The operations

Closure, interior, boundary

Distributive failures

Types of points

Key takeaways

The unit ball, from diamond to squareinteractivefree

2. Compactness: open covers, sequential, Heine–Borel

Compactness — the implication map free

In plain words

Why (0,1) is not compact — an explicit cover with no finite subcoverinteractivefree

3. Completeness and Baire category

Completeness, Banach fixed point and Baire category free

In plain words

Completeness

Banach contraction principle

Baire category theorem

Key takeaways

Baire category, and the argument it powersinteractive

4. Connectedness and path-connectedness

Connectedness — tests, preservation and the sine curve free

In plain words

Definitions

Quick facts

Standard examples

R\mathbb{R}R vs Rn\mathbb{R}^{n}Rn

Key takeaways

Connected but not path-connectedinteractive

$\mathbb{R}$ vs $\mathbb{R}^{n}$