On Definition 2.3 (F_del and G_sig): uncountable intersection example: irrat = bigcap { { r }^c : r a rational } are a countable intersection of open sets Open question (for us): is there any set that is neither G_del nor F_sig? 2.2 Neighborhoods and closures A^circ means "interior of A" means largest open set within A. --------X-----------| _______ | | . | V -------- - every set has a boundary but the boundary might be the empty set impose standard topology over X = [-4, 10] WRONG: boundary of X = { -4, 10 } and X is clopen Errattum: Your space X always has empty boundary as otherwise you'd have a point that non-trivially intersects with stuff outside the space, but by definition such stuff does not exist. We KNOW all open sets DO NOT CONTAIN their boundary points. Refer also to: https://www.amazon.com/Counterexamples-Topology-Dover-Books-Mathematics/dp/048668735X/ref=asc_df_048668735X/?tag=hyprod-20&linkCode=df0&hvadid=312165853622&hvpos=&hvnetw=g&hvrand=8743399671892861203&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9030244&hvtargid=pla-449277484053&psc=1&tag=&ref=&adgrpid=60258872297&hvpone=&hvptwo=&hvadid=312165853622&hvpos=&hvnetw=g&hvrand=8743399671892861203&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9030244&hvtargid=pla-449277484053 aka the counterexamples in topology book. REGARDING perfect subspaces: Consider the real line. { 4 } U [0, 1] is not perfect as shown by considering (3.5, 4.5) if 4 were a limit point in A there would exist a sequence, in A, converging to 4, containing stuff other than 4 implying that there are elements of A arbitrarily close to 4 implying that every neighborhood of 4 in A contains a point of A other than 4 namely one of those arbitrarily close things from the sequence { 4 - 1/n | n in Nat } will contain 4 - 1/2 = 3.5 but there is no sequence WITHIN that sequence which conveges to 3.5 hence 3.5 is not an accumulation point hence, the sequence is not perfect ( ( A B ) C ) in this one, ( A B ) is perfect because every neighborhood of A has B and vice versa so in conclusion they don't need to be "big" BTW If you ever have someone ask for a good intro to pre-orders and such things,. chapter 1 of this text is very very nice (according to me, Max): https://www.logicmatters.net/resources/pdfs/Galois.pdf a in A b in B (a, b) in A x B Munkres writes a x b in A x B a x b x c is in A x B x C as opposed to ((a, b), c) or (a, (b, c)) (a, b) in A x B and c in C (a, b) x c = (a, b, c) looks kind of weird compared to a x b x c also kind of nice in power-set spaces which you do a lot of in topo. so basically the notation handles the obvious isomorphism for you. { (ra, rb) | ra, rb are rational } (-pi, e) Union (-1 * first N digits of pi | N in nat, similar for e) Hence, we proved (ish) that R1 is 2nd countable. C(X) is the space of continuous functions X -> X A function f : X -> Y on topological spaces (X, T_X) and (Y, T_Y) is considered continuous if and only if, for all O_Y in T_Y, f^-1 O_Y is in T_X. ------ 2.2.7 Consider the topology (R, T) where T is defined below. T = { A | x in A -> exists U open in R, exists C countable, such that U - C is a subset of A } A base for T is the set B = { U - C where U is open in R and C is countable } Only points in this space that converge are ones that are eventually constant. 0----- Continued on 17 July 22 Suppose { x_alpha } alpha_in_A is a net. Let phi : Delta -> A be a function such that condition 2 holds. Then { x_phi(d) } d_in_Delta is called a subnet of the original net on line 3. phi Delta ----------> A n A --------------> X (1) Delta ----> A -----> X <------ d |----------------> x a such that n(x) = x a < a' -> f(a) < f(a') a in A then exists x in codomain(f) such that b > a -> { ()} ---- Cafe August 7 2022 ---- 2.28 Theorem - Let f : X->Y be a function and x in X. Let (X, T) and (Y, G) be the corresponding spaces. TFAE: I. f is continuous at x. II. if a net x_a -> x in X, then f(x_a) -> f(x) in Y. III. if a filter F -> x in X, then f(F) -> f(x) in Y. PROOF. Suppose f is continuous at x. <-> For every open neighborhood V of f(x), f^-1(V) is a neighborhood in X containing x. Let F be a filter in X and suppose x is a limit point of F. Therefore the closure of any s in F contains x. Thus f(closure(s)) contains f(x) for each s in F. Recall that f(closure(s)) <= closure(f(s)) as f is continuous. Thus closure(f(s)) contains f(x) for each s in F. But this suffices to show that f(F) -> f(x) in Y. So we have proven that I -> III. Suppose {x_a} is a net and x is a limit point of it. Consider the filter F generated by this net. It has the same limit points. So x is a limit point of F. So by prior logic, f(F) -> f(x) in Y. We still need to show f(x_a) -> f(x) in Y. NTS: Let V be an arbitrary nbrd of f(x), and a be arbitrary index. We claim: exists b >= a s.t. x_b in V. TODO - work on this more next time. 21st is next meeting September 4 2022 --- Proof of 2.32 Lemma in Hitchhiker's Guide: K is a compact subset of a T2 space Consider J = { (O, V) : k is in K and O is an open neighborhood of k and V is an open neighborhood of x and O intersect V = emptyset } Notice all these pairs must exist because the space is T2 Clearly the set of all the Os is an open covering of K By compactness of K, there exists a finite subset of this set, which is likewise an open covering of K Take the finite subset of corresponding Vs, and take its (finite) intsersection; this must also be open as finite intersection of open sets is open, and must include x because each V contains x Call this new set V_final Call the union of the finite collection of Os we came up with, O_final Clearly O_final intersect V_final = emptyset and O_final contains K and V_final contains x QED Claim: Compact subsets of T2 spaces are closed. recall that arbitrary intersection of closed sets is closed. Let K be a compact subset of a T2 space. For each point x not in K, let (O_x, V_x) be a tuple such that O_x, V_x are both open sets, O_x contains K, and V_x contains x, and the intersection of O_x, V_x is the empty set. Now take the intersection of all the V_x^c's for all the xs not in K. This is an arbitrary intersection of closed sets that must contain all of K. Indeed, this set cannot contain any x NOT in K, as to do so, it would need to contain an open neighborhood V_x, which cannot be the case as it's part of the intersection with V_x^c. So, this set doesn't contain anything not in K, and does contain K, therefore by excluded middle, the set is K. QED. Ansatz: A homomorphism of groups can be thought of as a set isomorphism. Let (G, +) and (H, *) be groups. Let f : G -> H be a homomorphism of groups. So, importantly, f(x + y) = f(x) * f(y). Consider the set G_zipped = { (x, y, x + y) : x, y in G } H_zipped = { () } Let h : G_zipped -> { (f(x), f(y), f(x + y)) | x, y in G } be the function h(x, y, x + y) = (f(x), f(y), f(x + y)) = (f(x), f(y), f(x) * f(y)) = H_zipped Because f is a homo. of groups, we were able to make a function h which was obviously equivalent, where h was a set isomorphism. f(a) = x if and only if (x, 0, x) = h(a, 0, a). G = (N, +) H = ({ 0 }, +) f(x) = 0 Conclusion: Max's Ansatz is wrong. ------------------------ 2.10 Semicontinuous Functions Let f : X -> R be a function. f is "lower semi-continuous" if for each r in R the pre-image of (-inf, r] is closed; and "upper semi-continuous" if for each r in R the pre-image of [r, inf) is closed. X = [1, 3] in R f (x) = x + 2 f(X) = [3, 5] which is not all of R minimizer of f is f^-1(3) = 1 [-inf, inf] -> [-inf, inf] minimizer = { -inf } {0, 1, 2} { x >= 3} f(x) = 0 if x <= 2 else 1 ( ( { . . . . . } ) { (.) ) . (.) . . . . . } R = (-inf, inf) [0, 1] = [-inf, inf] Suppose for a contradiction x is a minimizer Recall f : X -> R is our funciton, x in X By supposition, f(x) is the smallest