proof of Urysohn’s lemma
Let . Since is countable, we can use induction (or recursive definition if you prefer) to define the sets . List the elements of is an infinite sequence in some way; let us assume that and are the first two elements of this sequence. Now, define (the complement of in ). Since is a closed set of contained in , by normality of we can choose an open set such that and .
In general, let denote the set consisting of the first rationals in our sequence. Suppose that is defined for all and
Let be the next rational number in the sequence. Consider . It is a finite subset of so it inherits the usual ordering of . In such a set, every element (other than the smallest or largest) has an immediate predecessor and successor. We know that is the smallest element and the largest of so cannot be either of these. Thus has an immediate predecessor and an immediate successor in . The sets and are already defined by the inductive hypothesis so using the normality of , there exists an open set of such that
We now show that (1) holds for every pair of elements in . If both elements are in , then (1) is true by the inductive hypothesis. If one is and the other , then if we have
and if we have
Thus (1) holds for ever pair of elements in and therefore by induction, is defined for all .
We have defined for all rationals in . Extend this definition to every rational by defining
Then it is easy to check that (1) still holds.
Finally we show that this function we have defined satisfies the conditions of lemma. If , then for all so equals the set of all nonnegative rationals and . If , then for so equals all the rationals greater than 1 and .
To show that is continuous, we first prove two smaller results:
Proof. If , then for all so contains all rationals greater than . Thus by definition of .
Proof. If , then for all so contains no rational less than . Thus .
Let . Then since , (b) implies that and since , (a) implies that . Hence .
Finally, let . Then , so by (a). Also, so and by (b). Thus
as desired. Therefore is continuous and we are done.
|Title||proof of Urysohn’s lemma|
|Date of creation||2013-03-22 13:09:23|
|Last modified on||2013-03-22 13:09:23|
|Last modified by||scanez (1021)|