The Tychonoff Theorem

Transcription

1 Requirement for Department of Mathematics Lovely Professional University Punjab, India February 7, 2014

2 Outline Ordered Sets Requirement for 1 Ordered Sets

3 Outline Ordered Sets Requirement for 1 Ordered Sets 2

4 Outline Ordered Sets Requirement for 1 Ordered Sets 2 3 Requirement for

5 Outline Ordered Sets Requirement for 1 Ordered Sets 2 3 Requirement for 4

6 Requirement for Maximal and Minimal Elements of Ordered Sets I Let (A, ) be partial ordered set then An element a A is called maximal element of A if there exists no element in A which strictly dominate a i.e. x a, for every comparable elements x A An element b A is called minimal element of A if there exists no element in A which strictly precedes b i.e. b x, for every comparable elements x A

7 Requirement for Maximal and Minimal Elements of Ordered Sets II Let A = {3, 4, 5, 8, 9} be ordered by x divides y. Since 1 / A, if p is a prime number then its division are ±p. It means all prime number of A, i.e. 3 and 5 are its minimal elements. In particular a divides 2a and other higher integral multiples of a. hence there exists no maximal element in A. Consider the power set P (A) of A. Let this power set be ordered by set inclusion relation. Then φ is the least elements as well as minimal element of A. A is the maximal element as well as greatest element of A.

8 Requirement for Maximal and Minimal Elements of Ordered Sets III 1 is the minimal element of N with usual order and there exists no maximal element in N with usual order.

9 Requirement for Largest and smallest members of family Let P (A) and P (B) are families of sets A and B respectively, then An element A P (A) is called largest member or greatest member of P (A) if any C P (A) C A If A exists, then it will be a unique element of P (A). An element B P (B) is called smallest or least member of P (B) if any D P (B) B D If B exists, then it will be a unique element of P (B).

10 Requirement for Maximal and Minimal element of family Let P be a family of sets. let this set be ordered by set inclusion relation. Then A set B is called a proper set of A if B A and B A And we say that B is properly contained in A. An element A in P is called a maximal element of P if it is not properly contained in any element of P. More precisely any B P B and B A B is a proper subset of A An element A 1 in P is called a minimal element of P if any B P A 1 is a proper subset of B

11 Nest Ordered Sets Requirement for Definition A family P of sets is called a nest it it is linearly ordered by set inclusion relation. The following have same meaning : Linear order, nest, tower, chain. Let P and P 1 be families of sets such that P 1 is subset of P, then P 1 is called a nest in P.

12 Order Complete Ordered Sets Requirement for Definition An ordered set (X, ) is called order complete if sup(a) exists for each subset A of X. For example The ordered sets (N, <) and (Q, <) are ordered complete.

13 Initial Segment Ordered Sets Requirement for Definition Let (A, ) be a totally ordered set. Let a be arbitrary element of A, then set A a = {b A: b < a} is called an initial segment of A determined by an element of a of A. As (N, <) is totally ordered set, then initial segment of N determined by the element 5 of N is given by N 5 = {n N: n < 5} = {1, 2, 3, 4}

14 Requirement for First and last element of Chain I Definition Let (A, ) be a chain, then An element a A is called the first element of A if A a = φ or a < x for all x A i.e. a proceeds every element in A. An element b A is called the least element of A if x < b for all x A i.e. b dominates strictly every element which belong to A. N 1 = φ. Hence 1 is the first element of N with usual order and there is no last element.

15 Requirement for First and last element of Chain II R x φ, for all x R. Hence R with usual order has no first element. Q and Z with usual order have no first as well last elements. Let the power set P (A) be ordered by defining A 1 < A 2 if A 1 A 2, then φ is the first element and A is the last element of the linearly ordered set (P (A), <).

16 Well Order Set I Ordered Sets Requirement for Definition A linearly ordered set (A, ) is called a well ordered set if every subset of A has s first element. N with usual order is well order. Q, R and Z with usual order are now well order. Every subset of well order set is well order set.

17 Ordered Sets Requirement for Theorem Let A be a set that is strictly partial ordered. If every simple ordered subset of A has an upper bound in A, then A has a maximal elements.

18 Requirement for Requirement for I Lemma Let X be a set. Let A be a collection of subsets of X having FIP. Then there is a collection D of subsets of X such that D contain A and D has the FIP and no collection of subsets of X that property contain D has this FIP i.e. D is maximal w.r.t FIP.

19 Requirement for Requirement for II Lemma Let X be a set; let D be a collection of subsets of X that is maximal w.r.t. the FIP, then 1 Any finite intersection of elements of D is an element of D. 2 If A is a subset of X that intersects every element of D, then A is an element of D.

20 Requirement for Theorem An arbitrary product of compact spaces is compact in the product topology.

21 Proof: Let Ordered Sets Requirement for X = α J X α where each space X α us compact. Let A be collection of subsets of X having the finite intersection property. We prove that the collection A A A is non-empty. Now Compactness of X follows.

22 Requirement for Applying Lemma I, choose a collection D of subsets of X such that D A and D is maximal with respect to the FIP. It will suffice to show that the intersection D D D is non-empty. Given α J, let π α : X X α be the projection map, as usual. Consider the collection {π α (D): D D} of subsets of X α. This collection has the FIP because D does. By compacteness of X α, we can for each α choose a point x α of X α such that x α π α (D) D D Let x be the point (x α ) α J of X. We shall show that x D for every D D; then our proof will finished.

23 Requirement for First we show that if π 1 β (U β) is any subbasis element (for the product topology on X)containing x, then π 1 (U β) intersects every element of D. The set U β is a neighborhood of x β in X β. Since x β π β (D) by definition, U β intersect π β (D) in some point π β (y), where y D. Then it follows that y π 1 β (U β) D. It follows from (2) of Lemma-II that every subbasis element containing x belong to D. And then it follows from (1) of Lemma-II that every basis element containing x belong to D. Since D has the FIP, this means that every basis element containing x intersect every element of D. Hence x D for every D D as desired. β