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Homework 2 Connor O’Dell (3A1). If F is the collection of all closed bounded subset of Rstd together with R itself, then F is the family of closed sets for a topology on R strictly weaker than the usual topology. Proof. The intersection of any number of closed and bounded sets is again closed and bound. The first fact is clear and the last fact is true as the intersection must be contained in any one of the bounded sets in the intersection. Similarly, any finite union of closed and bounded sets is again closed and bounded. R is in the family by assumption, and ∅ is both closed and bounded. Thus F is a family of closed sets and defines a topology on R by Theorem 3.4. The open sets of this topology are a subset of the topology on Rstd , as theSare all the complements of closed sets in Rstd . Note that n∈N (n, n + 1) cannot be in this new topology, as its complement N is not bounded, but is open in Rstd . (3A4). Show that the collection of radially open sets is a topology for R2 . Compare this with Rstd . First note that ∅ is open vacuously, and R is radially open as it contains S all line segments. Suppose Ui is radially opend for i ∈ I. Let U = i∈I Ui . Let x ∈ U . Then x ∈ Uj for some j ∈ I. Thus there is an open line segment in each direction about x contained in Ui . These are also contained in U , so U is radially open. Suppose V1 and V2 are radially open. If y ∈ V1 ∩ V2 , there exist line segments in each direction about y in both V1 and V2 . For each direction about y, the shortest of these two line segments must be in both V1 and V2 . Thus V1 ∩ V2 is open. Induction shows that any finite intersection of open sets is open. Let τ be the set of radially open sets of R2 . We can see that τ is finer than Rstd . This is true as all open balls are also radially open, which implies that all open sets of Rstd are also open in τ . But τ has more open sets than Rstd as we can take the unit ball at zero and append both axes and still have an open set. 1 MATH 5610 Connor O’Dell (3A5). If A ⊂ X and τ is any topology for X, then {U ∪ (V ∩ A) | U, V ∈ τ } is a topology for X. Proof. Let σ = {U ∪ (V ∩ A) | U, V ∈ τ }. Suppose Si ∈ σ for i ∈ I. Then [ [ S= Si = (Ui ∪ (Vi ∩ A)) for some Ui , Vi ∈ τ i∈I i∈I = [ Ui [ ∪ i∈I = [ (Vi ∩ A) i∈I Ui [ ∪ i∈I Vi ∩ A i∈I which is in σ as τ must be closed under arbitrary unions. If S1 , S2 ∈ σ, S1 ∩ S1 = (U1 ∪ (V1 ∩ A) ∩ (U2 ∪ (V2 ∩ A) = ((U1 ∪ (V1 ∩ A)) ∩ U2 ) ∪ ((U1 ∪ (V1 ∩ A)) ∩ (V2 ∩ A)) = (U1 ∩ U2 ) ∪ (V1 ∩ U2 ∩ A) ∪ (U1 ∩ V2 ∩ A) ∪ (V1 ∩ V2 ∩ A) = (U1 ∩ U2 ) ∪ [(V1 ∩ U2 ) ∪ (U1 ∩ V2 ) ∪ (V1 ∩ V2 )] ∩ A which is in σ as τ is closed under finite intersections. We also see that ∅ = ∅ ∪ (∅ ∩ A) and X = X ∪ (X ∩ A), so ∅, X ∈ σ. (3D1). The complement of a regularly open set is regularly closed and vice versa. Proof. Suppose G is regularly open. Then in particular, Gc is closed and int(C c ) ⊆ Gc . Suppose F is a closed set containing int(Gc ). We show that Gc ⊆ F , and hence Gc is the smallest closed set containing int(Gc ) and so equal to int(Gc ) ⊆ Gc . If Gc * F , ∃x ∈ Gc such that x ∈ / F . Since F is closed (and F c is open), there exists an open set U such that x ∈ U ⊆ F c ⊆ G. thus U ∩ Gc = ∅, a contradiction of the fact that Gc is closed. Hence Gc ⊆ F . Now suppose G is regularly closed. Then Gc is open and Gc ⊆ int(Gc ), as Gc ⊆ Gc . We show that any open set contained in Gc is also contained in Gc , and hence int(Gc ) ⊆ Gc . Suppose V ⊆ Gc is open. Recall that Gc = Gc ∪ ∂Gc . If V * Gc , ∃x ∈ ∂Gc = ∂G ⊆ G. The last containment follows from the fact that G is closed. Since G is regularly closed, x must be in the interior of G or the closure of the interior of G. The first case is clearly impossible. In the second case, there must be some y ∈ V such that y ∈ int(G). Then there exists 2 MATH 5610 Connor O’Dell some open set A ⊆ G with y ∈ A. Now y ∈ A ∩ V ⊆ V , and A ∩ V is open. But V * Gc as y cannot be in Gc . Therefore we must have V ⊆ Gc . (3D2). There are open sets in R which are not regularly open. S Proof. U = n∈Z (n, n + 1) is open in Rstd and int(U ) = int(R) = R which is clearly not equal to U . (3D3). If A is any subset of a topological space, then int(cl(A)) is regularly open. Proof. int(cl(int(cl(A)))) = int(cl(int(A ∪ ∂A))) = int(cl((A ∪ ∂A) ∩ (∂(A ∪ ∂A)c ))) = int(cl((A ∪ ∂A) ∩ (∂A)c )) = int(cl((A ∪ ∂A) ∩ (int(A) ∪ int(Ac )))) = int(cl(int(A))) = int(cl(A)). (3D4). The intersection but not necessarily the union of two regularly open sets is regularly open. Proof. We see that (−∞, 0) and (0, ∞) are both regularly open in Rstd but the interior of the closure of their union is R. So the union of two regularly open sets is not necessarily regularly open. Let W = int(U ∩ V ). If U and V are regularly open, then U ∩ V is open and contained in U ∩ V , so U ∩ V ⊆ W . For the reverse containment, we first note the X is open and contained in U and V . So we also seet that W ⊆ int(U ) = U and W ⊆ int(V ) = V . Thus W ⊆ U ∩ V . (3F2). If ρ generates the topology on a metrizable space X and, for each λ ∈ Λ, Cλ isSa closed set in X such that ρ(Cλ1 , Cλ2 ) ≥ for all λ1 , λ2 , and > 0, then λ∈Λ Cλ is closed. S Proof. Let C = λ∈Λ Cλ . Let x ∈ C, and fix > 0. Then B(x, ) ∩ C 6= ∅. But any element of B(x, ) can only belong to one Cλ , so x ∈ Cλ = Cλ ⊆ C. Hence C = C, and C is closed. 3