Download TIETZE AND URYSOHN 1. Urysohn and Tietze Theorem 1. (Tietze

TIETZE AND URYSOHN
PETE L. CLARK
1. Urysohn and Tietze
Theorem 1. (Tietze Extension Theorem)
For a topological space X, the following are equivalent:
(i) X is quasi-normal.
(ii) If A ⊂ X is closed and f : A → [0, 1] is continuous, then there is a continuous
map F : X → [0, 1] with F |A = f .
(iii) For all disjoint closed subsets B1 , B2 of X, there is a Urysohn function: a
continuous function f : X → [0, 1] with B1 ⊂ f −1 (0) and B2 ⊂ f −1 (1).
Proof. (i) =⇒ (ii): We directly follow an argument of M. Mandelkern [Ma93]. Let
A ⊂ X be a closed subset of a quasi-normal topological space, and let f : A → [0, 1]
be a continuous function. For r ∈ Q, we put
Ar = f −1 ([0, r]),
so Ar ⊂ X is closed. For s ∈ Q ∩ (0, 1), we put
Us = X \ (A ∩ f −1 ([s, 1])),
so Us ⊂ X is open. Let
P = {(r, s) | r, s ∈ Q, 0 ≤ r < s < 1}.
The set P is countably infinite; let P = {(rn , sn )}∞
n=1 be an enumeration.
Let n ∈ Z+ . Inductively, we suppose that for all 1 ≤ k < n we have defined closed
subsets Hk ⊂ X such that
(1)
Ark ⊂ Hk◦ ⊂ HK ⊂ Usk ∀k < n
and
(2)
Hj ⊂ Hk◦ when j, k < n, rj < rk and sj < sk .
We will define Hn . First put
J = {j | j < n, rj < rn and sj < sn }
and
K = {k | k < n, rn < rk and sn < sk }.
Since X is quasi-normal, there is a closed subset Hn ⊂ X such that
∪
∩
Arn ∪
Hj ⊂ Hn◦ ⊂ Hn ⊂ Usn ∩
Hk◦ .
j∈J
k∈K
We write Hrs for Hn when r = rn and s = sn . Inductively, we have defined a
family {H(r,s) {(r,s)∈P of closed subsets of X such that
(3)
◦
∀(r, s) ∈ P, Ar ⊂ Hrs
⊂ Hrs ⊂ Us ,
1
2
PETE L. CLARK
◦
Hrs ⊂ Htu
when r < t and s < u.
(4)
For r ∈ Q ∩ [0, 1], put
Xr =
∩
Hrs .
s>r
For r < 0, let Xr = ∅. For r ≥ 1, let Xr = X. For (r, s) ∈ P , choose t ∈ Q such
that r < t < s. Then
∩
◦
Hsu = Xs .
Xr ⊂ Hrt ⊂ Hts
⊂ Hts ⊂
u>s
For r ∈ Q ∩ [0, 1), we have
Ar ⊂ Xr ∩ A = A ∩
∩
Hrs ⊂ A ∩
s>r
∩
Us = Ar .
s>r
Thus we have constructed a family {Xr }r∈Q of closed subsets of X such that
(5)
Xr ⊂ Xs◦ when r, s ∈ Q and r < s,
(6)
∀r ∈ Q, Xr ∩ A = Ar .
Finally, for x ∈ X put g(x) = inf{r | x ∈ Xr }. Then g : X → [0, 1]; since for all
x ∈ A we have f (x) = inf{r | x ∈ Ar }, we have that g|A = f . If a < b ∈ R then
∪
g −1 ((a, b)) = {Xs◦ \ Xr : r, s ∈ Q and a < r < s < b}
is open. Thus g is a continuous extension of f .
⨿
(ii) =⇒ (iii): Let B1 , B2 ⊂ X be closed and disjoint; put A = B1 ∪ B2 = B1 B2 .
The function g : A → [0, 1] with g|B1 ≡ 0 and g|B2 ≡ 1 is locally constant, hence
continuous. By assumption it extends to a continuous function f : X → [0, 1].
(iii) =⇒ (i): Let B1 , B2 ⊂ X be closed and disjoint. By our hypothesis, there
is a continuous function f : X → [0, 1] with f (B1 ) = {0}, f (B2 ) = {1}, let
U1 = f −1 ([0, 21 )), U2 = f −1 ( 12 , 1]). Then U1 , U2 ⊂ X are disjoint and open with
U1 ⊃ B1 and U2 ⊃ B2 .
Corollary 2. (Urysohn’s Lemma)
Normal spaces are Tychonoff. In particular compact spaces, regular Lindelöf spaces
and order spaces are Tychonoff.
Proof. Normal spaces are Hausdorff, so {p} is closed for all p ∈ X. So according to
Theorem 1 we can separate points from closed sets by continuous functions.
The following variant of Theorem 1 is also useful.
Corollary 3. Let X be quasi-normal, let A ⊂ X be closed, and let f : A → R be
continuous. Then there is a continuous map F : A → R such that F |A = f .
Proof. The obvious idea is the following: R is homeomorphic to (0, 1), so we may
as well asume that f (A) ⊂ (0, 1). Then in particular f (A) ⊂ [0, 1], so by TietzeUrysohn we may extend to a continuous function F : X → [0, 1]. However, this is
not good enough, since we don’t want F to take the values 0 or 1. (I.e.: we can
extend f : A ⊂ R to a continuous function to the extended real line [−∞, ∞].)
We get around this as follows: first, for shallow reasons to be seen shortly, it
will be better to work with the interval (−1, 1) instead of R. Certainly Theorem 1
TIETZE AND URYSOHN
3
holds for functions with values in [−1, 1] in place of [0, 1], so let F : X → [−1, 1]
such that F |A = f . Put
B = F −1 (0) ∪ F −1 (1),
so B ⊂ X is closed. Since F extends f and f (A) ∈ (0, 1), we have A ∩ B = ∅. Let
φ : X → [0, 1] be a Urysohn function for B and A: φ(B) = {0}, φ(A) = {1}. Put
h : X → [0, 1], h(x) = F (x)φ(x).
This works: h is a continuous extension of f with values in (−1, 1).
A subset of a topological space is a Gδ -set if it is a countable intersection of open
sets. A subset A of a topological space X is a zero set if there is a continuous
function f : X → [0, 1] with A = f −1 (0). If so, then
∞
∞
∩
∩
1
1
A = f −1 (
f −1 ([0, )),
[0, )) =
n
n
n=1
n=1
so A is a closed Gδ -set.
Let A, B be disjoint closed subsets in a topological space X. We have seen that
if X is quasi-normal, it admits a Urysohn function, i.e., a continuous function
f : X → [0, 1] with A ⊂ f −1 (0) and B ⊂ f −1 (1). It is natural to ask whether
we can always find a Urysohn function with A = f −1 (0) and B = f −1 (1): let us
call such an f a perfect Urysohn function for A and B and say that X is
perfectly normal if it is Hausdorff and a perfect Urysohn function exists for all
pairs of disjoint closed subsets.
Proposition 4. Metrizable spaces are perfectly normal.
Proof. Exercise!
Theorem 5. For a topological space X, the following are equivalent:
(i) X is perfectly normal.
(ii) X is separated and every closed subset of X is a zero set.
(iii) X is normal and every closed subset is a Gδ -set.
Proof. Exercise!
Proposition 6. If X is perfectly normal, then every subspace of X is normal.
Proof. Exercise!
References
[Ma93]
M. Mandelkern, A short proof of the Tietze-Urysohn extension theorem. Arch. Math.
(Basel) 60 (1993), 364-366.