202A: Solutions to Homework V Piotr Achinger Grading policy. In problem 3 (graded out of 4 points), part (e) was worth 2 points, part (d) was worth 1 point, and the remaining parts altogether were worth 1 point. Solutions Problem 1 (3 points). We start with an auxiliary result: Lemma. A space Y is compact if and only if, for every space X, the projection map π : X × Y −→ X is closed. Proof. (⇒) Let Z ⊆ X × Y be a closed subset, and let x ∈ X \ π(Z). As Z is closed, for every y ∈ Y , there exist opens x ∈ Uy ⊆ X, y ∈ Vy ⊆ Y such that (Uy × Vy ) ∩ Z = ∅. As y ∈ Y varies, the Vy form an open cover of Y . Since Y is compact, there exist y1 , . . . , yn ∈ Y such that Y = Vy1 ∪ . . . ∪ Vyn . Let U = Uy1 ∩ . . . ∩ Uyn ; as the intersection is finite, this is an open subset of X. Moreover, U ∩ π(Z) = ∅: if x0 ∈ U ∩ π(Z), then (x0 , y) ∈ U × Z for some y; let k be such that y ∈ Vyk , then (x0 , y) ∈ Uyk × Vyk , which is disjoint from Z, a contradiction. As x ∈ X \ π(Z) varies, these U form an open cover of X \ π(Z), hence π(Z) is closed. (⇐) Left to the reader (we will not need it here). Back to the problem; denote by πX : X × Y −→ X, πY : X × Y −→ Y the projections. Let W ⊆ Y be a closed subset; then πY−1 (Y ) is closed as well (as πY is continuous). Since Gr(f ) is closed too, Z := πY−1 (W ) ∩ Gr(f ) is closed. Then f −1 (W ) = πX (πY−1 (W ) ∩ Gr(f )) is closed by the Lemma, which shows that f is continuous. Problem 2 (3 points). (a) Let I ∗ = inf λ∈Λ Iλ∗ (f ). Note that I ∗ > −∞ as each Iλ∗ (f ) ≥ (b − a) · inf x∈[a,b] f (x). We claim that the net {Iλ∗ (f )}λ∈Λ converges to I ∗ . Let ε > 0. By the definition of I ∗ , there exists a λ0 ∈ Λ (depending on ε) such that Iλ∗0 (f ) < I ∗ + ε. If λ > λ0 , then Iλ∗ (f ) ≤ Iλ∗0 (f ), hence |Iλ∗ (f ) − I ∗ | < ε. (b) Analogous to (a). (c) Let λ0 = {a, c, b}. Then for every λ > λ0 , we have Iλ∗ (f ) = c − a = I∗λ (f ), therefore the limits are also equal. (d) As every interval contains both rational and irrational numbers, for every λ ∈ Λ we have Iλ∗ (f ) = 1 and I∗λ (f ) = 0. Thus the limits cannot be equal. Problem 3 (4 points). (a) Let A, B ∈ F , that is, A, B ⊆ X are such that {xλ }λ∈Λ is eventually in A and eventually in B. This means that there exist λ1 , λ2 ∈ Λ such that xλ ∈ Afor λ > λ1 , and xλ ∈ B for λ > λ2 . In particular, A and B are non-empty. As Λ is directed, there exists a λ3 ∈ Λ, λ3 > λ1 , λ2 . Then xλ ∈ A ∩ B for λ > λ3 , hence A ∩ B ∈ F , which proves that i) holds. Condition ii) clearly holds as well. 1 2 (b) Let A, B ∈ N (x). Since x ∈ A, B, they are both non-empty. By definition, there exist open subsets U ⊆ A, V ⊆ B containing x. Then A ∩ B contains U ∩ V , which is open and contains x, hence A ∩ B ∈ N (x), i.e., condition i) holds. Condition ii) clearly holds as well. (c) A net {xλ } converges to x if and only if for every open subset U ⊆ X containing x, {xλ } is eventually in U . This is equivalent to saying that every open U containing x is in the filter F defined in (a). If this is the case, as F satisfies condition ii), every subset of X containing an open neighborhood of x has to be in F as well, thus N (x) ⊆ F . Conversely, if N (x) ⊆ F , then for every open neighborhood U of x, we have U ∈ F , hence the net is eventually in U ; thus {xλ } converges to x. (d) Let F be a filter on X, and let F be the set of all F , ordered S filters on X containing S by inclusion.SLet C be a chain in F . We S check that C ∈ F ; clearly C ⊇ F , and the elements of C are non-empty. If A, B ∈ F , then A ∈ F1 , B ∈ F2 for some F1 , F2 ∈ C; as C is linearly ordered, we S must haveSF1 ⊆ F2 or F2 ⊆ F1 , in any case A ∩ B belongs to the bigger one, hence to C. If A ∈ C, A ⊆ B, then A ∈ F1 for some F1 ∈ C, hence S B ∈ F1 , thus B ∈ C as well. We have shown that every chain in F is bounded. By Zorn’s Lemma, there exists a maximal element F 0 of F , i.e., an ultrafilter containing F . (e) Let F be an ultrafilter on a set X, and let A ⊆ X. Suppose that A ∈ / F . Consider the family of subsets of X F 0 = {B ⊆ X : B ⊇ Ac ∩ F for some F ∈ F }. We claim that F 0 is a filter containing F . If B ⊇ Ac ∩ F for some F ∈ F , then B 6= ∅, for otherwise Ac ∩ F = ∅, hence F ⊆ A, which would imply that A ∈ F by condition ii). If B 0 ⊇ Ac ∩ F 0 for some F 0 ∈ F , then B ∩ B 0 ⊇ Ac ∩ (F 0 ∩ F ), which is in F 0 as F 0 ∩ F ∈ F . That condition ii) holds for F 0 and that F ⊆ F 0 is clear. Since F is an ultrafilter and F 0 ⊇ F , we must have F = F 0 . But Ac ∈ F 0 , so Ac ∈ F , as desired. (f) Suppose that {x} is closed but not open. One ultrafilter containing N (x) is the filter Fx consisting of all subsets containing x. Clearly, it contains N (x). Another example: take a net {xλ }λ∈Λ with limit x, such that xλ 6= x for all λ ∈ Λ (it exists as {x} is not open, i.e., x is not an isolated point of X). Consider the filter F corresponding to this net as in (a). By (c), we have N (x) ⊆ F . By part (d), there is an ultrafilter F 0 containing F . I claim that F 0 6= Fx . Indeed, the set U := X \ {x} is open (as by assumption {x} is closed), and we have U ∈ / Fx , but U ∈ F (as xλ ∈ U for all λ ∈ Λ). (g) In other words, I is the set of all subsets of N whose complements are finite. i): Since N is infinite, these are non-empty. The intersection of two such subsets has the union of two finite sets as a complement, which is finite. ii): If a subset has finite complement, any set containing it has a finite complement as well.