Topology

Introduction — Chapter 1, Lecture 1

NPTEL · Topology

Examples of Topological Spaces — Chapter 1, Lecture 2

NPTEL · Topology

The standard example zoo: discrete, indiscrete, cofinite.

Product Topology — Chapter 2, Lecture 6

NPTEL · Topology

Box and Product Topologies — Chapter 2, Lecture 8

NPTEL · Topology

Exactly where box and product differ on infinite products.

Topologies, bases and the standard constructions free

In plain words

A topology is a decision about which sets count as open. The surprise is the infinite product, where the useful topology is the one restricting all but finitely many coordinates.

Bases and comparison

A collection ℬ is a base iff it covers X and for $B_{1}, B_{2} \in$ ℬ and $x \in B_{1} \cap B_{2}$ there is $B_{3} \in$ ℬ with $x \in B_{3} \subseteq B_{1} \cap B_{2}$ .

$\tau_{1}$ is finer than $\tau_{2}$ if $\tau_{1} \supseteq \tau_{2}$ . Discrete is the finest, indiscrete the coarsest.

Topology on $\mathbb{R}$	Finer than usual?
discrete	yes, strictly
Sorgenfrey (lower limit, [a,b))	yes, strictly
usual	—
cofinite	no, strictly coarser
indiscrete	strictly coarser

The constructions

Subspace: open sets are U ∩ A.
Product (Tychonoff): base $= \prod{}U_{i}$ with $U_{i} = X_{i}$ for all but finitely many i. The box topology drops "finitely many" and is strictly finer on infinite products.
Quotient: U open in X/∼ iff its preimage is open. Quotients need not be Hausdorff even if X is.

Why the product topology wins: Tychonoff's theorem (arbitrary products of compact spaces are compact) is false for the box topology, and projections are continuous and open for the product topology.

Closure, interior, boundary

cl(A) = A ∪ A′; A is closed ⇔ A = cl A; A is dense ⇔ cl A = X.
int(A ∩ B) = int A ∩ int B, but int(A ∪ B) ⊇ int A ∪ int B can be strict $(A = \mathbb{Q}, B = \mathbb{R}$ ∖ $\mathbb{Q})$ .
cl(A ∪ B) = cl A ∪ cl B, but cl(A ∩ B) ⊆ cl A ∩ cl B can be strict (same example).

Key takeaways

Product ≠ box on infinite products, and Tychonoff needs the product topology.
Closure distributes over finite unions, interior over finite intersections — not the other way.
Quotients destroy Hausdorff; subspaces and products preserve it.

Which properties survive subspaces, products and quotientsinteractive

This visual is part of Trap Analysis

A table you can reconstruct — and the operation that destroys Hausdorff.

Exam focus: Compare topologies (finer/coarser) and know how the product topology differs from the box topology on infinite products — that difference is examined directly.

Practice

Part Cshow ▾

Let X be the "line with two origins": two disjoint copies

\mathbb{R}_{1}, \mathbb{R}_{2}

\mathbb{R}

, with

x \in \mathbb{R}_{1}

glued to

x \in \mathbb{R}_{2}

for every x ≠ 0, giving the quotient topology (the two origins

0_{1}, 0_{2}

stay distinct). Which of the following are true?

A.Every point of X has a neighbourhood homeomorphic to an open interval of $\mathbb{R}$ .✓ Correct
B.X is Hausdorff.
C.Every convergent sequence in X has a unique limit.
D.X is connected.✓ Correct

Solution

(1) each copy of $\mathbb{R}$ maps homeomorphically onto a neighbourhood of its points, including each origin.

(4) the images of $\mathbb{R}_{1}$ and $\mathbb{R}_{2}$ are both connected and share a point (e.g. the image of 1), so their union X is connected.

(2) is false: any neighbourhood of $0_{1}$ and any neighbourhood of $0_{2}$ both contain points x ≠ 0 arbitrarily close to 0, and those points are the same points in X — the neighbourhoods can never be disjoint.

(3) is false as a direct consequence: the sequence 1/n converges to both $0_{1}$ and $0_{2}$ at once.

Part Cshow ▾

Let X be a set and B a collection of subsets of X. Which of the following are true?

A.B is a basis for some topology on X iff every point of X lies in some element of B, and whenever $x \in B_{1} \cap B_{2}$ for $B_{1}, B_{2} \in B$ , there is $B_{3} \in B$ with $x \in B_{3} \subseteq B_{1} \cap B_{2}$ .✓ Correct
B.In the subspace topology on Y ⊆ X, the open sets of Y are exactly U ∩ Y for U open in X.✓ Correct
C.The basis {U × V : U open in X, V open in Y} for the product topology on X × Y consists precisely of the open sets of the product.
D.The quotient topology on X/∼ is defined so that a set is open iff its preimage under the quotient map is open in X.✓ Correct

Solution

(1) is the standard basis criterion.

(2) is the definition of the subspace topology.

(4) is the definition of the quotient topology.

(3) is false: {U × V} is a basis for the product topology, but a basis is a generating set, not the whole topology — arbitrary unions of such rectangles are also open, and most open sets in the product (an open disc in $\mathbb{R}^{2}$ , say) are not themselves a single rectangle U × V.

Part Bshow ▾

In the cofinite topology on an infinite set X, which is true?

A.X is Hausdorff
B.X is compact✓ Correct
C.Every subset is closed
D.X is discrete

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

Part Cshow ▾

Which of the following are true for topological spaces?

A.A product of two connected spaces is connected.✓ Correct
B.Every quotient of a Hausdorff space is Hausdorff.
C.Every subspace of a Hausdorff space is Hausdorff.✓ Correct
D.The product topology on $\mathbb{R} \times \mathbb{R}$ coincides with the usual Euclidean topology on $\mathbb{R}^{2}$ .✓ Correct

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

Part Cshow ▾

Which of the following are true?

A.cl(A ∪ B) = cl(A) ∪ cl(B) for any two subsets✓ Correct
B.cl(A ∩ B) = cl(A) ∩ cl(B) for any two subsets
C.int(A ∩ B) = int(A) ∩ int(B)✓ Correct
D.The box topology on $\mathbb{R}^\mathbb{N}$ is finer than the product topology✓ Correct

Solution

(b) fails for $A = \mathbb{Q}, B = \mathbb{R}$ ∖ $\mathbb{Q}$ : the left side is ∅, the right side is $\mathbb{R}$ .

2. Continuity, homeomorphism, separation axioms

Continuity and Related Concepts — Chapter 1, Lecture 5

NPTEL · Topology

T1 – Spaces and Hausdorff Spaces — Chapter 1, Lecture 4

NPTEL · Topology

Separation Axioms — Chapter 5, Lecture 19

NPTEL · Topology

Regular and Normal Spaces — Chapter 5, Lecture 20

NPTEL · Topology

Normality is the badly-behaved axiom — note which constructions preserve it.

Separation axioms and what survives which construction free

In plain words

Continuous means preimages of open sets are open. A homeomorphism needs the inverse continuous as well, which is not automatic — wrapping an interval onto a circle tears it.

The hierarchy

Axiom	Means
$T_{0}$	distinct points are topologically distinguishable
$T_{1}$	points are closed
$T_{2} ($ Hausdorff)	distinct points have disjoint neighbourhoods
$T_{3} ($ regular $+ T_{1})$	point and closed set separated
$T_{4} ($ normal $+ T_{1})$	disjoint closed sets separated

$T_{4} \Rightarrow T_{3} \Rightarrow T_{2} \Rightarrow T_{1} \Rightarrow T_{0}$ , all strict. Metric $\Rightarrow T_{4}$ , and compact Hausdorff $\Rightarrow T_{4}$ .

Inheritance table

Property	Subspace	Product	Continuous image	Quotient
Hausdorff	✓	✓	✗	✗
compact	closed subspaces only	✓ (Tychonoff)	✓	✓
connected	✗	✓	✓	✓
normal $(T_{4})$	closed subspaces only	✗ (Sorgenfrey plane)	✗	✗
metrizable	✓	countable products	✗	✗
separable	✗ in general (✓ for metric)	countable products	✓	✓

Handy facts

In a Hausdorff space, compact sets are closed and limits are unique.
A continuous bijection from a compact space to a Hausdorff space is a homeomorphism — the fastest way to prove a map is a homeomorphism.
Two continuous maps into a Hausdorff space agreeing on a dense set are equal.
Urysohn's lemma (normal spaces) gives separating functions; Urysohn metrization: second countable $+ T_{3} \Rightarrow$ metrizable.

Distinguishing spaces

Removing a point disconnects $\mathbb{R}$ but not $\mathbb{R}^{n} (n \ge 2)$ or $S^{1}$ , so no two of these are homeomorphic. [0,1] is compact, (0,1) is not.

Key takeaways

Compact + Hausdorff is the sweet spot: normal, and continuous bijections out of it are homeomorphisms.
Normality is the badly-behaved axiom — it fails for products and general subspaces.
Connectedness is not hereditary; compactness passes to closed subspaces only.

The separation axioms, in the order they matterinteractive

This visual is part of Trap Analysis

T₁ versus T₂ decided by one example, and the finite-space collapse that is a favourite question.

Exam focus: Know the implication chain T₄ ⇒ T₃ ⇒ T₂ ⇒ T₁ ⇒ T₀ (with the regularity/normality convention) and which properties are hereditary or productive.

Practice

Part C2025 Decembershow ▾

Let

\tau_E

denote the euclidean topology on

\mathbb{R}

. Which of the following statements are necessarily true?

A. $(\mathbb{R}, \tau_E)$ is a normal space.✓ Correct
B.Let $\tau$ be a topology on $\mathbb{R}$ . If the identity function of $\mathbb{R}$ is a continuous map from $(\mathbb{R},\tau_E)$ to $(\mathbb{R},\tau)$ , then $(\mathbb{R},\tau)$ is a regular space.
C.Let $\tau$ be a topology on $\mathbb{R}$ . If the identity function of $\mathbb{R}$ is a continuous map from $(\mathbb{R},\tau)$ to $(\mathbb{R},\tau_E)$ , then $(\mathbb{R},\tau)$ is a Hausdorff space.✓ Correct
D. $\mathbb{R}$ is a regular space in the finite-complement topology.

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

Part B2025 Juneshow ▾

Let S = {1, 2, 3, 4, 5} be equipped with the topology

\tau =

{∅, {1}, S}. What is the number of homeomorphisms of S onto itself?

A.25
B.120
C.24✓ Correct
D.6

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

Part C2024 Decembershow ▾

Let A, B, C be topological spaces such that A is homeomorphic to B, B is a subspace of C and the closure of B equals C. Let C be homeomorphic to a subspace W of A. Which of the following statements are FALSE?

A.The spaces B, the closure of W, and C are homeomorphic.✓ Correct
B.The spaces B, W, C are homeomorphic.✓ Correct
C.If C is compact, then A, B, C are homeomorphic.✓ Correct
D.If A is connected, then B and C are connected.

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

Part C2024 Decembershow ▾

Let X and Y be topological spaces and f : X → Y be a continuous function. Which of the following statements are true?

A.If X and Y are compact, Y is Hausdorff and f is onto, then X is also Hausdorff.
B.If X is an infinite compact set and f is a homeomorphism from X to f(X) (where f(X) is given the subspace topology), then Y is compact.
C.If X is Hausdorff and f is onto, then Y is Hausdorff.
D.If f is a homeomorphism, then X is second-countable if and only if Y is second-countable.✓ Correct

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

Part C2023 Decembershow ▾

Consider

\mathbb{R}^{2}

with the Euclidean topology and

\mathbb{Q}^{2} \subset \mathbb{R}^{2}

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

A. $\mathbb{Q}^{2}$ is connected.
B.If A is a non-empty connected subset of $\mathbb{Q}^{2}$ , then A has exactly one element.✓ Correct
C. $\mathbb{Q}^{2}$ is Hausdorff.✓ Correct
D.{ $(x, y) \in \mathbb{Q}^{2}$ | $x^{2} + y^{2} = 1$ } is compact in the subspace topology.

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

Part Bshow ▾

Every finite topological space satisfying the

T_{1}

axiom is

A.discrete✓ Correct
B.indiscrete
C.connected whenever it has more than one point
D.necessarily infinite

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

3. Compactness and Tychonoff

Compact Spaces — Chapter 4, Lecture 15

NPTEL · Topology

Properties of Compact Spaces — Chapter 4, Lecture 16

NPTEL · Topology

Compact ⊆ Hausdorff ⇒ closed, and continuous images stay compact.

Compactness in general spaces free

In plain words

Compactness still means every cover has a finite one hiding inside it, but outside metric spaces it no longer matches sequences. Tychonoff says any product of compact spaces is compact.

Definitions and equivalences

Compact = every open cover has a finite subcover = every family of closed sets with the finite intersection property has non-empty intersection.

In metric spaces: compact ⇔ sequentially compact ⇔ complete + totally bounded ⇔ countably compact. In general topological spaces these all differ.

What compactness gives

Continuous image of compact is compact ⇒ extreme value theorem.
Compact subset of a Hausdorff space is closed; closed subset of a compact space is compact.
Continuous bijection compact → Hausdorff is a homeomorphism.
Continuous f on a compact metric space is uniformly continuous (Heine–Cantor).
Tychonoff: any product of compact spaces is compact (equivalent to the axiom of choice).

Standard examples

Space	Compact?
$[a, b] \subset \mathbb{R}$	✓
closed unit ball in $\ell^{2}$ or C[0,1]	✗
$\mathbb{R}$ with the cofinite topology	✓
any infinite discrete space	✗
Cantor set	✓
{0} ∪ {1/n}	✓
$\mathbb{Q} \cap [0, 1]$	✗
one-point compactification of $\mathbb{R} \cong S^{1}$	✓

Riesz: the closed unit ball of a normed space is compact ⇔ the space is finite-dimensional.

Local compactness

$\mathbb{R}^{n}$ is locally compact; infinite-dimensional normed spaces and $\mathbb{Q}$ are not. Locally compact Hausdorff spaces admit a one-point compactification.

Key takeaways

Closed + bounded ⇒ compact only in $\mathbb{R}^{n}$ .
In metric spaces, compact ⇔ complete + totally bounded.
Compactness of the unit ball characterises finite dimension.

The compactness facts that need Hausdorff, and the ones that do notinteractive

This visual is part of Trap Analysis

Compact implies closed is false without it — here is the space where it fails.

Exam focus: Heine–Borel is ℝⁿ-only. In general spaces use open covers or the finite intersection property; in metric spaces sequential compactness is equivalent.

Practice

Part B2025 Decembershow ▾

Let X and Y be topological spaces. Consider the following statement: S: For every open subset U ⊆ X × Y and every x ∈ X such that {x} × Y ⊆ U, there is a neighbourhood W of x in X such that W × Y ⊆ U. Which of the following statements is true?

A.If X × Y is Hausdorff, then the statement S is true.
B.If Y is connected, then the statement S is true.
C.If X × Y is regular, then the statement S is true.
D.If Y is compact, then the statement S is true.✓ Correct

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

Part C2025 Decembershow ▾

Consider

X = \prod_{c\in(0,\infty)}

[0,c] together with the product topology, where [0,c] is equipped with the euclidean topology. Which of the following statements are necessarily true?

A.X is metrizable.
B.X is compact.✓ Correct
C.X is Hausdorff.✓ Correct
D.X is connected.✓ Correct

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

Part Cshow ▾

Which of the following statements about compact spaces are true?

A.A continuous bijection from a compact space onto a Hausdorff space is a homeomorphism.✓ Correct
B.A closed subspace of a compact space is compact.✓ Correct
C.A compact subspace of any topological space is closed in it.
D. $[0, 1]^\mathbb{N}$ with the product topology is compact.✓ Correct

Solution

(1) Closed subsets of the compact domain are compact, their images are compact and hence closed in the Hausdorff codomain — so the inverse map is continuous.

(2) A closed subset inherits the finite subcover property.

(3) needs Hausdorff: in the cofinite topology on $\mathbb{Z}$ , every subset is compact, including non-closed ones.

(4) is Tychonoff's theorem.

Part Bshow ▾

Which of the following subsets of

\mathbb{R} (

usual topology) is compact?

A.(0, 1]
B.{ $1/n : n \in \mathbb{N}$ }
C.{0} ∪ { $1/n : n \in \mathbb{N}$ }✓ Correct
D. $\mathbb{Q} \cap [0, 1]$

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

Part Cshow ▾

Let {

X_{i}

}

_{i\in{}I}

be a family of nonempty topological spaces. Which of the following are true?

A.By Tychonoff's theorem, the product $\prod{}X_{i}$ with the product topology is compact if every $X_{i}$ is compact.✓ Correct
B.Tychonoff's theorem for arbitrary index sets is equivalent to the Axiom of Choice.✓ Correct
C.The box topology on an infinite product of compact spaces is always compact.
D.The product topology is the coarsest topology on $\prod{}X_{i}$ making every projection map continuous.✓ Correct

Solution

(1) is Tychonoff's theorem.

(2) is the well-known equivalence (Kelley).

(4) is the defining universal property of the product topology.

(3) is false: Tychonoff's theorem is stated for the product topology specifically, and fails for the box topology — {0,1} $^\mathbb{N}$ with the box topology, for instance, is not compact, even though {0,1} is finite.

Part Cshow ▾

Let X be a topological space. Which of the following are true?

A.A continuous image of a compact space is compact.✓ Correct
B.A closed subset of a compact space is compact.✓ Correct
C.Every compact subset of a Hausdorff space is closed.✓ Correct
D.Every compact space is Hausdorff.

Solution

(1)–(3) are the standard compactness theorems.

(4) is false: compactness says nothing about separating points.

The cofinite topology on an infinite set X is compact (every open cover has a member missing only finitely many points, which a finite subcollection then mops up) but never Hausdorff (any two nonempty open sets, each missing only finitely many points, must intersect).

4. Connectedness and components

Connected Spaces — Chapter 3, Lecture 10

NPTEL · Topology

Pathwise Connected Spaces — Chapter 3, Lecture 13

NPTEL · Topology

Path-connected ⇒ connected; the converse is the topologist's sine curve.

Components and Introduction to Compact Spaces — Chapter 4, Lecture 14

NPTEL · Topology

Connectedness, components and totally disconnected spaces free

In plain words

Components are the largest connected chunks. They are always closed and not always open: in the rationals every component is a single point, and single points there are not open.

Components

The component of x is the union of all connected sets containing x. Components are closed, pairwise disjoint, and partition the space. They are open iff the space is locally connected.

In $\mathbb{R}$ , components of an open set are open intervals — hence an open subset of $\mathbb{R}$ is a countable disjoint union of intervals.
In $\mathbb{Q} ($ or the Cantor set, or $\mathbb{R}$ ∖ $\mathbb{Q})$ every component is a single point: totally disconnected. Here components are closed but not open.

Path components

Path components ⊆ components, with equality for locally path-connected spaces (e.g. open subsets of $\mathbb{R}^{n}$ , manifolds). Path components need not be closed — in the topologist's sine curve the graph piece is a non-closed path component.

Test table

Space	Connected	Components
$\mathbb{Q}$	✗	singletons
GL $_n(\mathbb{R})$	✗	2 (sign of det)
GL $_n(\mathbb{C})$ , SL $_n(\mathbb{R})$	✓	1
O(n)	✗	2
$\mathbb{R}$ ∖{0}	✗	2
$\mathbb{R}^{n}$ ∖{0}, n ≥ 2	✓	1
Cantor set	✗	singletons
cofinite topology on an infinite set	✓	1
topologist's sine curve	✓	1 (but 2 path components)

Uses

Continuous image of connected is connected — this is the IVT and the reason a connected space cannot surject continuously onto {0, 1}. $- \mathbb{R}$ and $\mathbb{R}^{n} (n \ge 2)$ are not homeomorphic: remove a point.
A connected space with more than one point and a clopen proper subset cannot exist — finding a clopen set is the standard disproof.

Key takeaways

Components are closed always, open only when the space is locally connected.
Totally disconnected ≠ discrete: $\mathbb{Q}$ is totally disconnected with no isolated points.
Path components can be non-closed even when components are.

The topologist's sine curve, walkedinteractive

This visual is part of Trap Analysis

Walk along sin(1/x) toward the segment and watch the distance meter diverge — connected, but no path gets there.

Exam focus: Components are always closed but not always open; in ℚ every component is a single point. Use clopen sets or a continuous surjection onto {0,1} to disprove connectedness.

Practice

Part B2024 Decembershow ▾

Which of the following statements is true?

A.{m + ne $^(2\pi{}i/3)$ | $m, n \in \mathbb{Z}$ } is a dense subset of $\mathbb{C}$ .
B.Open connected subsets of $\mathbb{R}^{3}$ need not be path-connected.
C.Let X be a topological space and $p : X \to \mathbb{R} a$ continuous surjective open map. If p⁻ $^{1}($ { $\alpha$ }) is connected for every $\alpha \in \mathbb{R}$ , then X must be connected.✓ Correct
D.Compact subsets of any infinite topological space are closed.

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

Part Cshow ▾

Let X be a topological space and, for x ∈ X, let C(x) denote the connected component of x. Which of the following are true?

A.C(x) is always closed in X.✓ Correct
B.C(x) is always open in X.
C.In a locally connected space, every connected component is open.✓ Correct
D.Two distinct connected components of X are always disjoint.✓ Correct

Solution

(1) the closure of a connected set is connected, so C(x) equals its own closure by maximality — always closed.

(3) is the standard theorem for locally connected spaces.

(4) components are the equivalence classes of "lies in a connected subset with", so distinct ones never meet.

(2) is false in general: in $\mathbb{Q}$ with the subspace topology, every component is a singleton, and singletons aren't open in $\mathbb{Q}$ .

Openness of components is a genuinely extra property, guaranteed only under local connectedness.

Part Cshow ▾

Which of the following spaces are connected?

A. $\mathbb{R}$ ∖ $\mathbb{Q}$
B.GL $_{2}(\mathbb{R})$
C.The topologist's sine curve✓ Correct
D.{ $(x, y) \in \mathbb{R}^{2}$ : xy = 1}

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

Part Cshow ▾

Which of the following are true about connectedness?

A.The continuous image of a connected space is connected.✓ Correct
B.The closure of a connected subspace is connected.✓ Correct
C.Every path-connected space is connected.✓ Correct
D.Every connected subset of $\mathbb{R}^{2}$ is path-connected.

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

Part Cshow ▾

Let X be a topological space. Which of the following are true?

A.If X is connected, then every continuous f : X → {0, 1} (discrete) is constant.✓ Correct
B.The union of two connected subsets with nonempty intersection is connected.✓ Correct
C.Every connected subset of $\mathbb{R} ($ usual topology) is an interval.✓ Correct
D.Path-connectedness and connectedness are equivalent for every topological space.

Solution

(1) is the standard characterisation of connectedness via maps to a two-point discrete space.

(2) and (3) are standard results.

(4) is false: the topologist's sine curve — the graph of sin(1/x) for x ∈ (0,1] together with the segment {0}×[−1,1] — is connected (it's the closure of a connected set) but not path-connected, since no path can reach the segment from the oscillating part.

The implication path-connected ⇒ connected always holds; the converse needs an extra hypothesis such as local path-connectedness.

Part Cshow ▾

Let

X = [0, 1] \cup [2, 3] \subseteq \mathbb{R}

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

A.X is disconnected.✓ Correct
B.X has exactly 2 connected components: [0,1] and [2,3].✓ Correct
C.[0,1] and [2,3] are both open and closed in X.✓ Correct
D.Every continuous function f : X → {0,1} (discrete) must be constant.

Solution

(1)–(3) follow from the gap between the two pieces: each piece is clopen in X, giving a genuine separation. (4) is false — it directly contradicts (1): a disconnected space always admits a non-constant continuous map to a two-point discrete space (send [0,1] to 0 and [2,3] to 1), which is exactly the criterion that fails here.

Part Cshow ▾

Let X be the continuous image of a connected space. Which of the following are true?

A.X is connected.✓ Correct
B.If f : Y → X is a continuous surjection and Y is connected, then X is connected.✓ Correct
C.A quotient space of a connected space is always connected.✓ Correct
D.If X is connected, then any space Y admitting a continuous surjection f : Y → X must also be connected.

Solution

(1)–(2) are the same standard fact stated twice. (3) a quotient map is a continuous surjection, so it follows immediately. (4) is false — it runs the implication backwards: connectedness of the image never forces connectedness of the domain. A constant map from the discrete two-point space {0,1} onto a single point {∗} is a continuous surjection whose codomain is (trivially) connected while its domain is not.

5. Standard spaces: cofinite, cocountable, Sorgenfrey, Cantor set

Examples of Topological Spaces — Chapter 1, Lecture 2

NPTEL · Topology

The counterexample zoo: discrete, indiscrete and cofinite topologies.

Interior Points, Limit Points — Chapter 1, Lecture 3

NPTEL · Topology

How closure and interior behave in these non-metric spaces.

The counterexample zoo: cofinite, cocountable, Sorgenfrey, Cantor free

In plain words

General topology has a small cast of stock examples, and most false claims are killed by one of them. Knowing which example breaks which claim beats reasoning from the definitions.

The property table

Property	Cofinite on $\mathbb{R}$	Cocountable on $\mathbb{R}$	Sorgenfrey $\mathbb{R}_\ell$	Cantor set
$T_{1}$	✓	✓	✓	✓
Hausdorff	✗	✗	✓	✓
compact	✓	✗	✗	✓
connected	✓	✓	✗	✗
separable	✓	✗	✓	✓
first countable	✗	✗	✓	✓
second countable	✗	✗	✗	✓
metrizable	✗	✗	✗	✓
normal	✓	✓	✓ (square is not)	✓

Cofinite topology

Open = ∅ or complement of a finite set. On an infinite set: $T_{1}$ but not Hausdorff, compact, connected, and every injective map into it is continuous. Convergence is strange: every sequence of distinct points converges to every point.

Sorgenfrey line $\mathbb{R}_\ell$

Base [a, b). Finer than the usual topology. Separable, first countable, Lindelöf, normal, totally disconnected — but not second countable and not metrizable. Its square is separable but contains an uncountable discrete subspace, so $\mathbb{R}_\ell^{2}$ is not normal: the standard counterexample to "product of normal is normal".

Cantor set

Compact, perfect (no isolated points), totally disconnected, uncountable, measure zero, nowhere dense, homeomorphic to {0,1} $^\mathbb{N}$ . Every compact metric totally disconnected perfect space is homeomorphic to it.

Key takeaways

Need "compact but not Hausdorff"? Cofinite. "Connected but not Hausdorff"? Cofinite.
Need "normal but square not normal"? Sorgenfrey.
Need "uncountable but measure zero"? Cantor.

The standard counterexample spaces, and what each is forinteractive

This visual is part of Trap Analysis

Cofinite, Sorgenfrey and the Cantor set — the three that answer most topology options.

Exam focus: These four spaces are the exam's stock counterexamples — memorise the property table and you can answer most 'which of the following is true' topology items by elimination.

Practice

Part C2024 Juneshow ▾

Let

\tau

be the smallest topology on the set

\mathbb{R}

containing

\beta =

{[a, b) |

a < b; a, b \in \mathbb{R}

}. Which of the following statements are true?

A. $\beta$ is a basis for topology $\tau$ ✓ Correct
B. $\mathbb{R}$ is compact in the topology $\tau$
C.Topology $\tau$ is the same as the Euclidean topology
D.Topology $\tau$ is Hausdorff✓ Correct

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

Part C2024 Juneshow ▾

Let X denote the topological space

\mathbb{R}

with the cofinite topology (i.e., the finite complement topology) and let Y denote the topological space

\mathbb{R}

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

A.X × [0, 1] is closed in X × Y with respect to the product topology✓ Correct
B.X × [0, 1] is compact with respect to the product topology✓ Correct
C.X is compact✓ Correct
D.X × Y is compact with respect to the product topology

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

Part C2024 Decembershow ▾

Define a topology

\tau

\mathbb{R}

as follows: a subset U of

\mathbb{R}

is in the topology

\tau

if and only if U = ∅ or 0 ∈ U. Which of the following statements are true?

A.The set of all irrational numbers is dense in $(\mathbb{R}, \tau)$ .
B.For each prime number p, the set { $0, \sqrt{p}$ } is dense in $(\mathbb{R}, \tau)$ .✓ Correct
C.[0, 1] is compact in $(\mathbb{R}, \tau)$ .
D. $(\mathbb{R}, \tau)$ is Hausdorff.

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

Part Bshow ▾

In the Sorgenfrey line

(\mathbb{R}

with basis the intervals [a, b)), the set [0, 1) is

A.open but not closed
B.closed but not open
C.both open and closed✓ Correct
D.neither open nor closed

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

Part Bshow ▾

The Cantor set is

A.countable and of measure zero
B.uncountable and of measure zero✓ Correct
C.uncountable and of positive measure
D.countable and nowhere dense

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

Part Cshow ▾

Let X be an infinite set with the cofinite topology. Which of the following are true?

A.X is compact✓ Correct
B.X is Hausdorff
C.X is connected✓ Correct
D.Every infinite subset of X is dense✓ Correct

Solution

Any two non-empty open sets meet (their complements are finite), so X is not Hausdorff but is connected; the closure of an infinite set is X.