Question

Question 3. Lul (X', d) be a metric space. A subsct ACX is said to be Gy if there exista a collection of open U u ch that A- , , Similarly, a subact BCis said to be F if there exista collection of closed sets {F}x=1 such that A=U- Fr. 1) Prove that A is G, if and only if A is T. b) Prove that every closed subset of (.X.d) is G- c) Prove that is not Gain (RID. (Hint: Baire Category Theorem) d) Prove that there does not exists a function /:R → R that is continuous al cach point in Q yet discontinuons at each point in R 0. (Hint: See Assignment 1, Question 7) Theorem (Baire's Category Theorem) Let (X.d) be a complete metric price. Suppose ( U R> is a sequence of open dense subels of X. Then Unis dense in X. Hence X is of second category in itself. Question 7. Let: R R and let DU)={IER f is discontinuous at s}. D. For cach n e N let for every > there exists y, 2 such that I 14 Ir-y<0,1-2|< , and f(y)-f(x) a) Show for cach n EN that Daf) is closed. b) Show that D()-U-D. ). Hence the discontinuities off is a countable union of closed sets. Definition Let(X, d) be a metric space. A subset ACX is said to be • nowhere dense if int(A) = 0. • first category in (X,d) if A=UA-1 A, where each An CX is nowhere dense. • second category in (X. d) if A is not first category. • residual il 4 is first category Definition The set A is called the closure of Definition Let ( d) be a metric space and let ACX. An element 30 EX is said to be a: • limit point of A if there exists a sequence (an> of points in A that converges to : • cluster point of A if there exists a sequence (47)x>1 of points in A that convergento 2 {ro} • boundary point of A ifo is a limit point of both A and A". • interior point of A if there exists an e > 0 such that Bt.) CA. The set of limit, cluster, boundary, and interior points of are denoted lim(A), cluster(A), bdy(A), and int(A) respectively. Before we get to examples, we note the following which will give us an alternative characterization of lim(A), cluster(A), and idy(A) which look more like the characterization of int(A). Lemma Let (X. d) be a metric space and let ASX. Then x e lim(A) if and only if An B(x,c) 7 for all > 0. Consequently rebdy(A) if and only is for alle > Have An B(1,) and An B(r.) 0. Similarly cluster(A) if and only if An Br.c)\{2} + for all c>0. Definition Let (24.d) be a metric space and let ASBS X. IL is said that is dense in B if BCA. Equivalently, A is dense in B if for every be B there exists a sequence (an) of elements from A such that Lemma Let (x,d) be a metric space and let ACX Then A is closed and nowhere dense if and only if 10 is open and dense in X. Corollary Let (X,d) be a metric space. Then X is of first category in itself if and only if there crists a sequence (Un)r>1 of open dense subsets of X with 0.= .

The question that is being asked is Question 3 that has a red rectangle around it.

The subsection on Question 7 is just for the Hint to part d of Question 3.

Answer 1

exists Question 3 Given: - let (x,d) be a melho space A subset ACX is said to be Gs if there a collection of open sets {un? such that A= A Un Similarly, a subset मा

| B n (a) Spin C X is said to be Fo it there exists a collection of closed sets {Foto such that A= U Fn Prove that A is in G. if and only if A info This follows directly from the de morgan laws A and CA Anl= u An no nal and the fact that complements of open sets are closed (and vice versa) (Here Y=xly, the complement of Y in x) CL To prove :- every closed Subset of (x, do is Gg. pobog: let nein consider Bon= F = U B(26) which is a collection of open ball. This is to conta Go set. If at n Bn then for each n by a the definition of Br there is some an EF with dian,a) 1 Then and hx} , so as f is closed and XEF Elence, it follows the result. (c) To show: - Q is not Gg in CIR, 1.1) proff: 94 Q were a Gg then the complement of Q is the countable union of closed sets Q = U n since Qe has no interior (n) = 0, where & denotes the interior of a set. But we've written R=Q UCUGA) as a countable union of nowhere deinse sets and this contradicts the Baire Category theorem