MATH 307 MIDTERM PROOFS

Nested open balls. Statement: If 0<a<b, then B(x,a)⊆B(x,b).

Proof: Let y∈B(x,a). Then d(x,y)<a<b, so y∈B(x,b). Hence B(x,a)⊆B(x,b).

Every open ball is open. Statement: For x∈ℝⁿ and r>0, B(x,r) is open.

Proof: Take y∈B(x,r) and ε=r−d(x,y)>0. If z∈B(y,ε), then d(x,z)≤d(x,y)+d(y,z)<d(x,y)+ε=r. Thus B(y,ε)⊆B(x,r), so B(x,r) is open.

Continuity and open preimages. Statement: f:ℝⁿ→ℝᵐ is continuous iff f⁻¹(U) is open for every open U⊆ℝᵐ.

Proof: (⇒) For x∈f⁻¹(U), choose ε>0 with B(f(x),ε)⊆U. Continuity gives δ>0 with B(x,δ)⊆f⁻¹(U); hence the preimage is open. (⇐) For x and ε>0, V=f⁻¹(B(f(x),ε)) is open and contains x. Choose δ>0 with B(x,δ)⊆V; then d(x,y)<δ ⇒ d(f(x),f(y))<ε.

A basis generates a topology. Statement: If ℬ satisfies the two basis axioms and 𝒯ℬ={U⊆X: ∀x∈U ∃B∈ℬ with x∈B⊆U}, then 𝒯ℬ is a topology.

Proof: ∅ qualifies vacuously; X qualifies by basis coverage. For x∈⋃Uα, choose Uα containing x, then B∈ℬ with x∈B⊆Uα⊆⋃Uα. For x∈U∩V, choose B₁⊆U and B₂⊆V containing x; the basis intersection axiom gives B₃∈ℬ with x∈B₃⊆B₁∩B₂⊆U∩V. Thus all topology axioms hold.

Open sets are unions of basis elements. Statement: For a basis ℬ, U is open in 𝒯ℬ iff U is a union of elements of ℬ.

Proof: (⇒) For each x∈U choose Bₓ∈ℬ with x∈Bₓ⊆U. Then U=⋃ₓ∈U Bₓ. (⇐) If U=⋃Bα, every x∈U lies in some Bα⊆U, so U∈𝒯ℬ.

Comparing topologies using bases. Statement: If ℬ₁,ℬ₂ are bases for 𝒯₁,𝒯₂ on X, then 𝒯₁⊆𝒯₂ iff ∀B₁∈ℬ₁, ∀x∈B₁, ∃B₂∈ℬ₂ with x∈B₂⊆B₁.

Proof: (⇒) B₁∈𝒯₁⊆𝒯₂, so the ℬ₂-basis gives B₂ with x∈B₂⊆B₁. (⇐) For U∈𝒯₁ and x∈U, choose B₁∈ℬ₁ with x∈B₁⊆U, then B₂∈ℬ₂ with x∈B₂⊆B₁⊆U. Hence U∈𝒯₂.

A subbasis generates a basis. Statement: If 𝒮 covers X, its finite intersections form a basis ℬ𝒮.

Proof: Coverage: each x lies in some S∈𝒮, which is a finite intersection. Intersection axiom: if x∈B₁∩B₂ with B₁,B₂∈ℬ𝒮, then B₁∩B₂ is again a finite intersection of members of 𝒮 and contains x. Thus ℬ𝒮 is a basis.

Topology generated by a subbasis. Statement: The topology generated by a subbasis 𝒮 consists of arbitrary unions of finite intersections of 𝒮-elements; it is the smallest topology containing 𝒮.

Proof: Finite intersections of 𝒮 form a basis (Prop. 2), so their arbitrary unions form a topology (Thm. 4). It contains 𝒮. Any topology containing 𝒮 contains its finite intersections and arbitrary unions, so this is the smallest such topology.

Open rectangles form a basis. Statement: The family {U×V: U open in X, V open in Y} is a basis on X×Y.

Proof: X×Y itself is an open rectangle, so the family covers X×Y. Also (U₁×V₁)∩(U₂×V₂)=(U₁∩U₂)×(V₁∩V₂), another open rectangle. Both basis axioms hold.

Recognizing an existing basis. Statement: If ℬ consists of open subsets of X and every open U and x∈U admit B∈ℬ with x∈B⊆U, then ℬ is a basis for the original topology 𝒯.

Proof: Taking U=X gives coverage. For x∈B₁∩B₂, the intersection is open, so the hypothesis gives B₃ with x∈B₃⊆B₁∩B₂. Thus ℬ is a basis. Since each B∈ℬ is 𝒯-open, 𝒯ℬ⊆𝒯; the hypothesis gives 𝒯⊆𝒯ℬ. Hence equality.

Product basis equals projection subbasis topology. Statement: On X×Y, the subbasis {U×Y: U open in X}∪{X×V: V open in Y} generates the usual open-rectangle product topology.

Proof: Each subbasic set is an open rectangle. Conversely U×V=(U×Y)∩(X×V), a finite intersection of subbasis elements. Thus each topology contains a basis for the other; they coincide.

Products of bases. Statement: If ℬ₁ and ℬ₂ are bases for X and Y, then {B₁×B₂: B₁∈ℬ₁, B₂∈ℬ₂} is a basis for X×Y.

Proof: Each B₁×B₂ is open in the product topology. For any open rectangle U×V containing (x,y), choose B₁∈ℬ₁ and B₂∈ℬ₂ with x∈B₁⊆U and y∈B₂⊆V. Then (x,y)∈B₁×B₂⊆U×V. By the basis-recognition lemma, these products form a basis.

Preimage of a composition. Statement: For f:X→Y, g:Y→Z, U⊆Z: (g∘f)⁻¹(U)=f⁻¹(g⁻¹(U)).

Proof: For x∈X: x∈(g∘f)⁻¹(U) ⇔ g(f(x))∈U ⇔ f(x)∈g⁻¹(U) ⇔ x∈f⁻¹(g⁻¹(U)).

Compositions are continuous. Statement: If f:X→Y and g:Y→Z are continuous, then g∘f:X→Z is continuous.

Proof: For any open U⊆Z, (g∘f)⁻¹(U)=f⁻¹(g⁻¹(U)). Continuity of g makes g⁻¹(U) open; continuity of f makes its preimage open. Hence g∘f is continuous.

Check continuity on a basis. Statement: If ℬ is a basis for Y, f:X→Y is continuous iff f⁻¹(B) is open for every B∈ℬ.

Proof: (⇒) Basis elements are open, so their preimages are open. (⇐) Every open U⊆Y is a union of basis elements. Thus f⁻¹(U)=⋃ f⁻¹(B), a union of open sets, so f is continuous.

Preimages preserve unions and intersections. Statement: For any family {Uα} of subsets of Y: f⁻¹(⋃Uα)=⋃f⁻¹(Uα) and f⁻¹(⋂Uα)=⋂f⁻¹(Uα).

Proof: For any x, f(x)∈⋃Uα iff ∃α with f(x)∈Uα, proving the union identity; f(x)∈⋂Uα iff ∀α, f(x)∈Uα, proving the intersection identity.

Check continuity on a subbasis. Statement: If 𝒮 is a subbasis for Y, f:X→Y is continuous iff f⁻¹(S) is open for every S∈𝒮.

Proof: (⇒) Subbasis elements are open, so their preimages are open. (⇐) Suppose every f⁻¹(S) is open. Every open set is a union of finite intersections of subbasis elements; preimages preserve both. Hence all open sets have open preimages, so f is continuous.

Continuity into a two-factor product. Statement: If f:W→X and g:W→Y are continuous, then (f,g):W→X×Y is continuous.

Proof: For open U⊆X and V⊆Y, (f,g)⁻¹(U×V)=f⁻¹(U)∩g⁻¹(V), which is open. Since open rectangles form a basis for X×Y, (f,g) is continuous.

Restriction of a continuous map. Statement: If f:X→Y is continuous and A⊆X, then f|A:A→Y is continuous (subspace topology).

Proof: For open U⊆Y, (f|A)⁻¹(U)=A∩f⁻¹(U). Since f⁻¹(U) is open in X, this intersection is open in A. Thus f|A is continuous.

Restricting the codomain. Statement: If f:X→Y is continuous and f(X)⊆B⊆Y, then f:X→B is continuous with B given the subspace topology.

Proof: For V open in B, write V=B∩U with U open in Y. Because f(X)⊆B, f⁻¹(V)=f⁻¹(B∩U)=f⁻¹(U), which is open in X. Thus f:X→B is continuous.

Properties of closed sets. Statement: In X, (i) ∅,X are closed; (ii) arbitrary intersections of closed sets are closed; (iii) finite unions of closed sets are closed.

Proof: (i) Their complements are open. (ii) X∖(⋂Fα)=⋃(X∖Fα), an open union. (iii) X∖(⋃ᵢ₌₁ⁿFᵢ)=⋂ᵢ₌₁ⁿ(X∖Fᵢ), a finite open intersection. Thus the respective sets are closed.

Closed sets in a subspace. Statement: For A⊆Y⊆X: A is closed in Y iff A=Y∩K for some K closed in X.

Proof: (⇒) Y∖A=Y∩U for U open in X; hence A=Y∩(X∖U), with X∖U closed. (⇐) If A=Y∩K and K closed, then Y∖A=Y∩(X∖K), open in Y. Thus A is closed in Y.

Closure is the smallest closed superset. Statement: For A⊆X, cl(A) is closed, contains A, and lies in every closed F⊇A.

Proof: By definition cl(A)=⋂{F: F is closed and A⊆F}. Arbitrary intersections of closed sets are closed; each F contains A, so cl(A) contains A; as an intersection, cl(A)⊆F for every such F.

Closure and limit points. Statement: If A′ is the set of limit points of A, then cl(A)=A∪A′.

Proof: (⊇) Any x∈A∪A′ has every neighborhood meeting A, so x∈cl(A). (⊆) If x∈cl(A), every neighborhood meets A. If x∈A, done; otherwise each neighborhood contains a point of A different from x, so x∈A′.

Closed sets contain their limit points. Statement: A⊆X is closed iff A′⊆A.

Proof: A closed ⇔ cl(A)=A ⇔ A∪A′=A (Thm. 16) ⇔ A′⊆A.

Uniqueness of sequence limits. Statement: In a Hausdorff space, a sequence has at most one limit.

Proof: Suppose xₙ→x and xₙ→y with x≠y. Choose disjoint open U∋x and V∋y. Eventually xₙ∈U and xₙ∈V simultaneously, contradicting U∩V=∅. Thus x=y.

Products of Hausdorff spaces. Statement: Any product ∏ⱼ∈J Xⱼ of Hausdorff spaces is Hausdorff in the product topology.

Proof: For distinct x,y in the product, choose j with xⱼ≠yⱼ. Hausdorffness of Xⱼ gives disjoint open U∋xⱼ and V∋yⱼ. Then πⱼ⁻¹(U) and πⱼ⁻¹(V) are disjoint open neighborhoods of x and y. Hence the product is Hausdorff.

Diagonal map fails box continuity. Statement: With the box topology on ℝ^ω, f:ℝ→ℝ^ω, f(t)=(t,t,…) is not continuous.

Proof: Let U=∏ₙ₌₁^∞(−1/n,1/n), open in the box topology. Then f⁻¹(U)=⋂ₙ₌₁^∞(−1/n,1/n)={0}, not open in ℝ. Thus f is not continuous.

Continuity into a countable product. Statement: For f:W→∏ᵢ₌₁^∞Xᵢ (product topology), f is continuous iff each πᵢ∘f is continuous.

Proof: (⇒) πᵢ∘f is continuous as a composition of continuous maps. (⇐) If each πᵢ∘f is continuous, then for U open in Xᵢ, f⁻¹(πᵢ⁻¹(U))=(πᵢ∘f)⁻¹(U) is open. These πᵢ⁻¹(U) form a subbasis, so f is continuous.

Coordinate projections are continuous. Statement: For a product ∏ᵢ∈J Xᵢ in the product topology, each projection πⱼ:∏Xᵢ→Xⱼ is continuous.

Proof: For any open U⊆Xⱼ, πⱼ⁻¹(U) is a subbasis element of the product topology, hence open. Thus πⱼ is continuous.

Continuity into an arbitrary product. Statement: If each fⱼ:W→Xⱼ is continuous and f(w)=(fⱼ(w))ⱼ∈J, then f:W→∏ⱼ∈J Xⱼ is continuous.

Proof: For U open in Xⱼ, f⁻¹(πⱼ⁻¹(U))=(πⱼ∘f)⁻¹(U)=fⱼ⁻¹(U), which is open. Since sets πⱼ⁻¹(U) form a subbasis, f is continuous.

Sequence Lemma. Statement: If X is metrizable and p is a limit point of A⊆X, there exists a sequence aₙ∈A with aₙ→p.

Proof: Choose a compatible metric d. For each n, pick aₙ∈A∩B(p,1/n) with aₙ≠p (possible since p is a limit point). Then d(aₙ,p)<1/n→0, so aₙ→p.

Uncountable product is not metrizable. Statement: If J is uncountable, then ∏ⱼ∈Jℝ with the product topology is not metrizable.

Proof: Let p=(0)ⱼ and A={x∈{0,1}ᴶ: x(j)=0 at only finitely many j}. Every basic neighborhood of p restricts finitely many coordinates, so meets A; thus p is a limit point. If aₙ∈A converged to p, the union F of their finite zero-coordinate sets would be countable. Pick j₀∈J∖F; then aₙ(j₀)=1 for all n, contradicting convergence to p under continuous πⱼ₀. By the Sequence Lemma, the product is not metrizable.

Connectedness and clopen sets. Statement: X is connected iff its only clopen subsets (both open and closed) are ∅ and X.

Proof: (⇒) A nonempty proper clopen A and X∖A would separate X, contradiction. (⇐) If X=U∪V were a separation, U would be nonempty, proper, and clopen, contradiction.

Separations via limit points. Statement: If X=A∪B with A,B disjoint nonempty, then (A,B) is a separation iff A∩B′=∅ and B∩A′=∅.

Proof: (⇒) In a separation A,B are closed, hence A′⊆A and B′⊆B; disjointness gives the two equalities. (⇐) Since X=A∪B, no cross-limit points implies A′⊆A and B′⊆B. Thus A,B are closed, hence (as complements) open; so they separate X.

Limit points in a subspace. Statement: If A⊆Y⊆X and p∈Y, then p is a limit point of A in Y iff it is a limit point of A in X.

Proof: Any neighborhood of p in Y has the form U∩Y with U open in X. As A⊆Y, (U∩Y)∩(A∖{p})=U∩(A∖{p}). Thus every neighborhood in Y meets A∖{p} iff every neighborhood in X does.

Connected subset cannot cross a separation. Statement: If X=U∪V with U,V disjoint open and C⊆X connected, then C⊆U or C⊆V.

Proof: If C met both, C∩U and C∩V would be disjoint, nonempty, open in C, and cover C; this would separate C, a contradiction.

Union of two connected sets. Statement: If A,B⊆X are connected and A∩B≠∅, then A∪B is connected.

Proof: Suppose A∪B=U∪V is a separation. As A,B are connected, neither can meet both sides; each lies entirely in U or V. Their nonempty intersection forces the same side, say U, so V=∅, contradiction.

Union of a family meeting a connected set. Statement: If C and each Cα are connected and C∩Cα≠∅ for every α, then C∪⋃αCα is connected.

Proof: Suppose the union has a separation U,V. Connected C lies on one side, say U. Each connected Cα meets C, so also lies entirely in U. Then V=∅, contradiction.

Continuous images preserve connectedness. Statement: If f:X→Y is continuous and X is connected, then f(X) is connected.

Proof: Suppose f(X)=U∪V is a separation. Then f⁻¹(U),f⁻¹(V) are disjoint, nonempty, open subsets of X covering X, contradicting connectedness. Hence f(X) is connected.

Product of connected spaces. Statement: If X and Y are connected, then X×Y is connected.

Proof: If X×Y=∅, done. Otherwise fix y₀∈Y. The horizontal slice A=X×{y₀} and each vertical slice Bₓ={x}×Y are connected (homeomorphic to X,Y), with A∩Bₓ={(x,y₀)}≠∅. By Thm. 26, X×Y=A∪⋃ₓ∈X Bₓ is connected.

Finite products of connected spaces. Statement: If X₁,…,Xₙ are connected, then ∏ᵢ₌₁ⁿ Xᵢ is connected.

Proof: Induct on n. Base n=1 is immediate. If ∏ᵢ₌₁ⁿ Xᵢ is connected and Xₙ₊₁ is connected, then (∏ᵢ₌₁ⁿ Xᵢ)×Xₙ₊₁ is connected by Thm. 28. Hence all finite products are connected.

Between connected set and its closure. Statement: If A⊆B⊆cl(A) and A is connected, then B is connected.

Proof: Suppose B=U∪V is a separation. Connected A lies in one side, say U. As U is closed in B and contains A, cl_B(A)⊆U. But B⊆cl_X(A) implies cl_B(A)=B, so B⊆U and V=∅, contradiction.

The unit interval is connected. Statement: [0,1] is connected in its usual topology.

Proof: Suppose [0,1]=U∪V is a separation, with 0∈U and b∈V. Set c=sup(U∩[0,b]). Since U is closed in [0,1], c∈U. Since b∈V, c<b. Since U is open, some x∈U satisfies c<x<b, contradicting that c is the supremum. Thus [0,1] is connected.

Neighborhoods in [0,1]. Statement: If x₀∈U⊆[0,1] and U is open in [0,1], then U contains a one-sided interval at x₀=0 or 1, and a two-sided interval for 0<x₀<1.

Proof: Write U=O∩[0,1], with O open in ℝ. Choose δ>0 with (x₀−δ,x₀+δ)⊆O. Taking sufficiently small ε>0 gives [0,ε)⊆U if x₀=0, (1−ε,1]⊆U if x₀=1, or (x₀−ε,x₀+ε)⊆U otherwise.

Continuity via closed sets. Statement: f:X→Y is continuous iff f⁻¹(F) is closed in X for every closed F⊆Y.

Proof: (⇒) If F is closed, Y∖F is open, so X∖f⁻¹(F)=f⁻¹(Y∖F) is open; hence f⁻¹(F) is closed. (⇐) If U is open, Y∖U is closed, so X∖f⁻¹(U)=f⁻¹(Y∖U) is closed; hence f⁻¹(U) is open and f continuous.

Pasting Lemma. Statement: If X=A∪B with A,B closed, and f|A,f|B are continuous, then f:X→Y is continuous.

Proof: For closed C⊆Y, (f|A)⁻¹(C) and (f|B)⁻¹(C) are closed in A,B, hence closed in X because A,B are closed. Their union is f⁻¹(C), so f⁻¹(C) is closed. Thus f is continuous.

Reparameterization of a path. Statement: If a<b and f:[a,b]→X is a path, then g(t)=f(a+t(b−a)) is a path [0,1]→X with the same endpoints.

Proof: Let h(t)=a+t(b−a). The map h is continuous; hence g=f∘h is continuous. Also g(0)=f(a) and g(1)=f(b).

Path relation is an equivalence relation. Statement: Define x∼y iff a path in X joins x and y. Then ∼ is an equivalence relation; its classes are path components.

Proof: Reflexive: t↦x is a constant path. Symmetric: reverse a path f by f(1−t). Transitive: if f joins x to y and g joins y to z, concatenate h(t)=f(2t) for 0≤t≤1/2 and h(t)=g(2t−1) for 1/2≤t≤1. The formulas agree at 1/2; the Pasting Lemma makes h continuous. Thus ∼ is an equivalence relation.

Path-connected implies connected. Statement: If X is path-connected, then X is connected (converse false in general).

Proof: Suppose X=U∪V is a separation, with x∈U and y∈V. Path-connectedness gives continuous f:[0,1]→X with f(0)=x and f(1)=y. Then f⁻¹(U),f⁻¹(V) are disjoint, nonempty, open, and cover [0,1], contradicting connectedness of [0,1]. Hence X is connected.