Countability Axioms

D 1 Countability axioms We have studied four basic countability properties: (1) The first countability axiom. (2) The second countability axiom, (3) (4) The Lindel6f condition. tse condition that the space has a countable dense subset. We know that condition (2) implies each of the others. We show now that this is the only-general theorem relating these four conditions. We shall in fact find, for each subset of conditions (1), (3), and (4), a space that satisfies the conditions in the subset, and none of the others. This requires seven distinct examples! Incidentally, there do exist relations among these four conditions for certain types of spaces. For instance, for metrizable spaces, condition (1) is automatically satisfied, and the other three conditions are equivalent to one another. (See Exercise 5 of 30.) Similarly, for topological groups that are first-countable, conditions (2), (3), and () are equivalent. (See.Exercise 18 of §30.) Example 1. Conditions (1), (3), and (4). The space JPR is first-countable, Lindelof, and has a countable dense subset , but is not second-countable. (See Example 3 of Conditions (1) and (3). Ex:ample 2. Lindel6f. 30.) The ordered square is compact, and hence It does not have It is readily seen to be first-countable. a countable dense subset, since each dense subset must contain at least one point of each interval ax (0,1). Conditions (1) and (4). DEample 3. The space IPjx [R and the rational points form a countable dense subset. is first-countable, It is not Lindelof; see Example 4 of §30. Conditions (3) and (4). Example 4. the proof given in Example 2 of §21 for The space II R by the Tychonoff theorem, so it is Lindelof. subset of I as follows: is not first-countable; works also here. It is compact, We construct a countable dense D.2 Given a partition 0 = a0 < al of the interval = 1 ... •an I = [0,1], where the ai are rational, and given a sequence bl,... ,bn of rational numbers, let us define a step function f(x) =b. for ~~(,"~ fa)) f(a Then f is an element of ai_. f: I--->I x _ ai by setting fi = 1,...,n) b = bn - n ) I ; and the set of all such f is countable. We shall show that these functions form a dense subset of BLt us take a typical basis element B f6r II; iI. it is the intersection of finitely many sets of the form -i c (Ui) 1 for i = 1,...,ni set of where I, for each i. c 1 < c 2 The set < ... cn B are points of I and Ui consists of all functions from is an open I to whose graphs intersect the vertical intervals in the following figure. I-1 -3 (A PI{ 3 Cl Given B, I I L Cs C 9 let us choose rational numbers 0 - aa~3 O0- Cl< 1 al< a 1 < cC,< 2 Then, for each i, choose b The corresponding function f a. 1 such that ... < Crrt a r. to be a rational number in the open set ha1 s the graph pictured; it consists of horizontal line segments with rational end points. ObT­ 1 eo 4 .T 6 C) e j a 1~~~ 1. l o - a2o 43 If U. I D.3 Example 5. not Lindel6f. Condition (1). The space SYL is first-countable, but it is (Take the open cover by sets of the form S , for *t< 2L .) Nor does it have a countale dense subset. Example 6. Condition (3). TI-le space does it have a countable dense subset. Example 7. Condition (4). Example 2 of §21. S, is not first-countable, nor But it is Lindelbf, being compact. The space RI is not first-countable; see Nor is it Lindel6f; for it is regular, and a regular Lindel6f space is normal (see Exercise 4 of §32); but (See Notes F.) The space (0,1) I , which is an open subspace of dense subset. is not normal. Finally, we note that if a space has a countable dense subset, then so does any open subspace of it. space RI II; R is homeomorphic to the tlerefore it has a countable