Topology: Bases, Constructions, Connectedness, Compactness and the Separation Axioms
1. Topologies, bases and subbases
A topology on X is a family of subsets containing ∅ and X, closed under arbitrary unions and finite intersections. A basis 𝓑 is a family covering X such that for B₁, B₂ ∈ 𝓑 and x ∈ B₁ ∩ B₂ some B₃ ∈ 𝓑 has x ∈ B₃ ⊆ B₁ ∩ B₂; the open sets are then the unions of basis elements. Any family 𝓢 covering X is a subbasis: its finite intersections form a basis. Example: on X = {a, b, c}, the subbasis {{a, b}, {b, c}} generates ∅, {b}, {a, b}, {b, c}, X — five open sets.
| Topology | Open sets | Profile |
|---|---|---|
| discrete | every subset | metrizable; totally disconnected; compact iff finite |
| indiscrete | ∅ and ℝ | connected, compact, not T₁ |
| cofinite | ∅ and complements of finite sets | T₁, not Hausdorff; connected; every subset compact |
| lower limit ℝℓ | unions of [a, b) | finer than usual; normal; first countable, separable, Lindelöf, not second countable, not metrizable; totally disconnected |
2. Subspace, order, product, quotient and metric topologies
- Subspace: the open sets of A ⊆ X are U ∩ A, U open in X. In A = [0, 1] ∪ (2, 3], the set [0, 1] is open (= A ∩ (−1, 3/2)) and closed.
- Order topology on a linearly ordered set: basis of open intervals and rays. On ℝ it is the usual topology; on ℤ it is discrete; the order topology on [0, 1] × [0, 1] in dictionary order is not the product topology.
- Product topology on Π Xₐ: generated by the sets πₐ⁻¹(U), so basic open sets restrict only finitely many coordinates. It is the coarsest topology making every projection continuous, and a map into the product is continuous iff each coordinate is. The box topology, restricting all coordinates, is finer and differs for infinite products.
- Quotient topology from a surjection q: X → Y: V is open iff q⁻¹(V) is open. [0, 1] with 0 ∼ 1 is the circle S¹; the square with opposite edges identified the same way is the torus, with one pair reversed the Klein bottle.
- Metric topology: basis of open balls. Metrizable spaces are Hausdorff, normal and first countable, so a space failing any of these is not metrizable.
3. Connectedness and path-connectedness
X is connected if its only clopen subsets are ∅ and X; path-connected if any two points are joined by a path. Path-connected ⇒ connected; connected and locally path-connected (e.g. an open subset of ℝⁿ) ⇒ path-connected. Continuous images, closures, and unions of connected sets with a common point are connected, and arbitrary products of connected spaces are connected in the product topology. The components are the maximal connected subsets; they are closed and partition X.
- The cofinite topology on an infinite set is connected: two non-empty open sets have finite complements and must meet.
- ℝℓ is totally disconnected: [a, ∞) is open and also closed, its complement (−∞, a) being a union of intervals [a − n, a).
- {(x, y) : xy ≠ 0} has 4 components, the open quadrants. GL₂(ℝ) has 2, det > 0 and det < 0, because det is continuous and never 0 on it and each sign class is path-connected.
4. Compactness, sequential and limit point compactness, Tychonoff
X is compact if every open cover has a finite subcover; sequentially compact if every sequence has a convergent subsequence; limit point compact if every infinite subset has a limit point. Compact ⇒ limit point compact always; for metric spaces the three are equivalent; in general neither of compact and sequentially compact implies the other.
| Fact | Hypothesis needed | Counterexample without it |
|---|---|---|
| a compact subset is closed | X Hausdorff | cofinite ℝ: [0, 1] is compact and not closed |
| a closed subset of a compact space is compact | none | — |
| a continuous bijection from a compact space is a homeomorphism | target Hausdorff | the identity from ([0, 1], usual) to ([0, 1], indiscrete) is a continuous bijection from a compact space and not a homeomorphism |
| compact ⇔ closed and bounded | subsets of ℝⁿ (Heine–Borel) | closed unit ball of ℓ² |
Tychonoff’s theorem: an arbitrary product of compact spaces is compact in the product topology (it is equivalent to the axiom of choice). So [0, 1]ℕ and {0, 1}ℕ (the Cantor set) are compact. In the box topology [0, 1]ℕ is not compact. The one-point compactification of ℝⁿ is Sⁿ.
5. Countability and separation axioms; Urysohn and Tietze
| Axiom | Meaning | Relations |
|---|---|---|
| first countable | countable neighbourhood base at each point | every metric space |
| second countable | countable basis | ⇒ first countable, separable and Lindelöf; for metric spaces ⇔ separable ⇔ Lindelöf |
| separable | countable dense subset | ℝℓ is separable (ℚ) but not second countable |
| Lindelöf | every open cover has a countable subcover | ℝℓ is Lindelöf; ℝℓ × ℝℓ is not |
Separation axioms. T₁: points are closed. T₂ (Hausdorff): distinct points have disjoint neighbourhoods. Regular (T₃): a point and a closed set not containing it have disjoint neighbourhoods, plus T₁. Normal (T₄): two disjoint closed sets have disjoint neighbourhoods, plus T₁. T₄ ⇒ T₃ ⇒ T₂ ⇒ T₁, each strictly: the cofinite topology on ℝ is T₁ and not Hausdorff. Metric spaces and compact Hausdorff spaces are normal; ℝℓ is normal but ℝℓ × ℝℓ is not.
- Urysohn’s lemma: X is normal iff for any disjoint closed A, B there is a continuous f: X → [0, 1] with f = 0 on A and f = 1 on B. In a metric space f = d(x, A)/(d(x, A) + d(x, B)) does it directly.
- Tietze extension theorem: if X is normal and A ⊆ X is closed, every continuous f: A → [a, b] (or → ℝ) extends to a continuous F: X → [a, b] (or ℝ). Closedness is essential: sin(1/x) on (0, 1] ⊂ ℝ has no continuous extension to ℝ.
- Urysohn metrization: every regular, second countable (T₁) space is metrizable.
Key takeaways
- A basis generates open sets by unions and a subbasis by finite intersections then unions; know the discrete, indiscrete, cofinite and lower-limit topologies by heart.
- Product topology restricts finitely many coordinates; quotients glue: [0, 1]/0 ∼ 1 is S¹ and the square with both pairs of edges identified is the torus.
- Path-connected ⇒ connected, not conversely; products of connected spaces are connected; cofinite ℝ is connected and ℝℓ totally disconnected.
- Compact subsets of Hausdorff spaces are closed; compact ⇒ limit point compact; Tychonoff makes products of compact spaces compact.
- T₄ ⇒ T₃ ⇒ T₂ ⇒ T₁; metric and compact Hausdorff spaces are normal; Urysohn separates closed sets by functions and Tietze extends functions from closed sets.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The number of distinct topologies on a set with two elements is ____.
Numerical answer — type the value.
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Answer: 4
On {a, b} a topology contains ∅ and X and any selection of {a}, {b} closed under ∪ and ∩: {∅, X}, {∅, {a}, X}, {∅, {b}, X}, and the discrete one. That is 4; only the last is Hausdorff.On X = {a, b, c}, the number of open sets in the topology generated by the subbasis {{a, b}, {b, c}} is ____.
Numerical answer — type the value.
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Answer: 5
Finite intersections of subbasis elements give the basis {a, b}, {b, c}, {a, b} ∩ {b, c} = {b}, together with X (the empty intersection). Unions then give ∅, {b}, {a, b}, {b, c}, X — 5 open sets. {a} and {c} are not open.Which of the following spaces are Hausdorff?
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Answer: B — every metric space; C — ℝ with the lower limit topology generated by [a, b)
(1) No: two non-empty open sets have finite complements and always meet. (2) Balls of radius d(x, y)/2 separate x and y. (3) Yes: it is finer than the usual topology, which is already Hausdorff. (4) No: the only neighbourhood of either point is the whole set.In ℝ with the lower limit topology (basis {[a, b) : a < b}), the set [0, 1) is
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Answer: A — both open and closed
[0, 1) is a basis element, so it is open. Its complement (−∞, 0) ∪ [1, ∞) = ⋃[−n, 0) ∪ ⋃[1, n) is a union of basis elements, so it is open and [0, 1) is closed. Such clopen sets are why the lower limit line is totally disconnected.The standard basis for the product topology on X × Y consists of
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Answer: A — all sets U × V with U open in X and V open in Y
Open rectangles U × V are closed under finite intersection, (U₁ × V₁) ∩ (U₂ × V₂) = (U₁ ∩ U₂) × (V₁ ∩ V₂), and cover X × Y, so they form a basis; unions of them are the product topology. The sets U × Y alone form only part of a subbasis.The quotient of the square [0, 1] × [0, 1] obtained by identifying (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y) for all x, y is homeomorphic to
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Answer: A — the torus S¹ × S¹
Identifying one pair of opposite edges with the same orientation gives a cylinder S¹ × [0, 1]; identifying its two boundary circles, again the same way, gives S¹ × S¹. Reversing one pair gives the Klein bottle.Which statements about connectedness are true?
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Answer: A — ℝ with the cofinite topology is connected; C — a product of connected spaces is connected in the product topology
(1) Any two non-empty open sets meet (cofinite complements), so no separation exists. (2) [0, ∞) is clopen. (3) holds for arbitrary products. (4) is false: no path joins a point of the vertical segment to the graph of sin(1/x), although the curve is connected.The number of connected components of {(x, y) ∈ ℝ² : xy ≠ 0} is ____.
Numerical answer — type the value.
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Answer: 4
Removing both axes leaves the four open quadrants. Each is convex, hence connected; and the continuous functions sign(x) and sign(y) are constant on components, so points in different quadrants lie in different components.The number of path components of GL₂(ℝ), the group of invertible real 2 × 2 matrices with the topology of ℝ⁴, is ____.
Numerical answer — type the value.
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Answer: 2
det is continuous and non-zero on GL₂(ℝ), so it cannot change sign along a path: det > 0 and det < 0 are separated. Each is path-connected (any matrix of positive determinant can be deformed to I, e.g. through Gram–Schmidt to a rotation). So there are 2.Which of the following spaces are compact?
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Answer: A — ℝ with the cofinite topology; B — [0, 1]^{ℕ} with the product topology
(1) Any one open set of a cover misses only finitely many points, each covered by one more set. (2) Tychonoff. (3) The cover (1/n, 1] has no finite subcover. (4) The singletons form an open cover with no finite subcover.Which statements are true for topological spaces?
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Answer: A — a compact subset of a Hausdorff space is closed; B — a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism; D — a closed subset of a compact space is compact
(1) For y ∉ K, Hausdorffness and compactness give finitely many neighbourhoods of K and one of y that are disjoint. (2) Closed sets map to compact, hence closed, sets, so the inverse is continuous. (4) Add the open complement to any cover. (3) is false: in the cofinite topology on ℝ, [0, 1] is compact and not closed.Which statements about the forms of compactness are true?
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Answer: A — in a metric space, compact and sequentially compact are equivalent; B — every compact space is limit point compact
(1) is the metric theorem. (2) An infinite set with no limit point is closed and discrete, so its points have neighbourhoods containing no other point; with the complement these cover X with no finite subcover. (3) fails: ℕ × {0, 1}, with ℕ discrete and {0, 1} indiscrete, is limit point compact but the open sets {n} × {0, 1} have no finite subcover. (4) ℤ ⊂ ℝ has no limit point.For the lower limit line ℝℓ (basis {[a, b)}), which statements are true?
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Answer: A — it is first countable; B — it is separable
(1) {[x, x + 1/n)} is a countable neighbourhood base at x. (2) ℚ meets every [a, b). (3) fails: any basis must contain, for each x, a set with minimum x inside [x, x + 1), giving uncountably many distinct sets. (4) fails: a separable metric space is second countable.Which statements about separation axioms are true?
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Answer: A — every metric space is normal; B — every compact Hausdorff space is normal; D — every regular second countable T₁ space is metrizable
(1) U = {d(x, A) < d(x, B)} and V = {d(x, B) < d(x, A)} separate disjoint closed A, B. (2) Compactness upgrades point-separation to set-separation twice. (4) is Urysohn’s metrization theorem. (3) fails: the cofinite topology on ℝ is T₁ and not Hausdorff.Urysohn’s lemma states that a T₁ space X is normal if and only if
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Answer: A — for any disjoint closed sets A and B there is a continuous f: X → [0, 1] with f = 0 on A and f = 1 on B
Given such f, f⁻¹[0, 1/2) and f⁻¹(1/2, 1] are disjoint open sets around A and B, so the functional condition implies normality; the lemma is the converse, built from a dyadic family of open sets. Normal spaces need not be metrizable (ℝℓ), and bounded functions characterise compactness-type properties, not normality.f(x) = sin(1/x) on A = (0, 1] has no continuous extension to ℝ. This does not contradict the Tietze extension theorem because
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Answer: A — A is not a closed subset of ℝ
Tietze extends continuous functions from closed subsets of a normal space. ℝ is metric, hence normal; f is continuous on A and bounded by 1. The failure is that 0 is a limit point of A outside A, where f oscillates, so A is not closed.