proof of Urysohn’s lemma
First we construct a family of open sets of indexed by the
rationals such that if , then . These
are the sets we will use to define our continuous function.
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
(1) |
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.
Now, given , define . This
set contains no number less than and contains every number greater
than by the definition of for and . Thus
is bounded below and its infimum is an element in
. Define
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:
(a)
Proof. If , then for all so contains all rationals greater than . Thus by definition of .
(b) .
Proof. If , then for all so contains no rational less than . Thus .
Let and let be an open interval of
containing . We will find a neighborhood of such that
. Choose such that
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 |
---|---|
Canonical name | ProofOfUrysohnsLemma |
Date of creation | 2013-03-22 13:09:23 |
Last modified on | 2013-03-22 13:09:23 |
Owner | scanez (1021) |
Last modified by | scanez (1021) |
Numerical id | 4 |
Author | scanez (1021) |
Entry type | Proof |
Classification | msc 54D15 |