Urysohn’s lemma which characterizes normal spaces is a very important result which is
useful in general topology and in the construction of measures. Because it is somewhat
technical a proof is given for the part which is needed.
Theorem 10.2.1(Urysohn) Let
(X, τ)
be normaland let H ⊆ U where H isclosed and U is open. Then there exists g : X → [0,1] such that g is continuous,g
(x)
= 1 on H and g
(x)
= 0 if x
∕∈
U.
Proof: Let D ≡{r_{n}}_{n=1}^{∞} be the rational numbers in (0,1]. Choose V_{r1} an open set
such that
--
H ⊆ Vr1 ⊆ V r1 ⊆ U.
This can be done by applying the assumption that X is normal to the disjoint closed sets,
H and U^{C}, to obtain open sets V and W with
H ⊆ V,UC ⊆ W, and V ∩ W = ∅.
Then
-- -- C
H ⊆ V ⊆ V ,V ∩U = ∅
and so let V_{r1} = V .
Suppose V_{r1},
⋅⋅⋅
,V_{rk} have been chosen and list the rational numbers r_{1},
If r_{k+1}> r_{lk} then letting p = r_{lk}, let V_{rk+1} satisfy
Vp ⊆ Vr ⊆ Vr ⊆ U.
k+1 k+1
If r_{k+1}∈ (r_{li},r_{li+1}), let p = r_{li} and let q = r_{li+1}. Then let V_{rk+1} satisfy
V- ⊆ V ⊆ V- ⊆ V .
p rk+1 rk+1 q
If r_{k+1}< r_{l1}, let p = r_{l1} and let V_{rk+1} satisfy
--
H ⊆ Vrk+1 ⊆ Vrk+1 ⊆ Vp.
Thus there exist open sets V_{r} for each r ∈ ℚ ∩
(0,1)
with the property that if
r < s,
-- --
H ⊆ Vr ⊆ Vr ⊆ Vs ⊆ Vs ⊆ U.
Now let
⋃
f (x) = min (inf{t ∈ D : x ∈ Vt},1),f (x) ≡ 1 if x ∕∈ Vt.
t∈D
(Recall D = ℚ ∩ (0,1].) I claim f is continuous.
f−1([0,a)) = ∪ {Vt : t < a,t ∈ D },
an open set.
Next consider x ∈ f^{−1}
([0,a])
so f
(x)
≤ a. If t > a, then x ∈ V_{t} because if not,
then
f (x) ≡ inf{t ∈ D : x ∈ Vt} > a.
Thus
f− 1([0,a]) ⊆ ∩{Vt : t > a} = ∩{V-t : t > a}
which is a closed set. If x ∈∩{V_{t} : t > a}, then x ∈∩{V_{t} : t > a} and so f
(x)
≤ a.
If a = 1,f^{−1}
([0,1])
= f^{−1}
([0,a])
= X. Therefore,
f− 1((a,1]) = X ∖f− 1([0,a]) = open set.
It follows f is continuous. Clearly f
(x)
= 0 on H. If x ∈ U^{C}, then x
∈∕
V_{t} for any t ∈ D so
f
(x )
= 1 on U^{C}. Let g
(x )
= 1 − f
(x)
. This proves the theorem.
In any metric space there is a much easier proof of the conclusion of Urysohn’s lemma
which applies.
Lemma 10.2.2Let S be a nonempty subset of a metric space,
(X,d)
. Define
f (x) ≡ dist(x,S) ≡ inf{d(x,y) : y ∈ S }.
Then f is continuous.
Proof: Consider
|f (x)− f (x )|
1
and suppose without loss of generality that
f
(x )
1
≥ f
(x)
. Then choose y ∈ S such that f
(x)
+ ε > d
(x,y)
. Then
|f (x1)− f (x)| = f (x1)− f (x ) ≤ f (x1)− d(x,y)+ ε
≤ d(x1,y)− d(x,y)+ ε
≤ d(x,x1)+ d(x,y)− d(x,y)+ ε
= d(x1,x)+ ε.
Since ε is arbitrary, it follows that
|f (x1) − f (x)|
≤ d
(x1,x)
and this proves the
lemma.
Theorem 10.2.3(Urysohn’s lemma for metric space) Let H be a closed subset ofan open set, U in a metric space,
(X,d)
. Then there exists a continuous function,g : X →
[0,1]
such that g
(x)
= 1 for all x ∈ H and g
(x)
= 0 for all x
∈∕
U.
Proof: If x
∕∈
C, a closed set, then dist
(x,C )
> 0 because if not, there would exist a
sequence of points of C converging to x and it would follow that x ∈ C. Therefore,
dist
(x,H)
+ dist
( )
x,UC
> 0 for all x ∈ X. Now define a continuous function, g
as
( )
g (x) ≡ -----distx,U-C-------.
dist(x,H )+ dist(x,UC )
It is easy to see this verifies the conclusions of the theorem and this proves the
theorem.
Theorem 10.2.4Every compact Hausdorff space is normal.
Proof:First it is shown that X, is regular. Let H be a closed set and let p
∈∕
H. Then
for each h ∈ H, there exists an open set U_{h} containing p and an open set V_{h} containing
h such that U_{h}∩V_{h} = ∅. Since H must be compact, it follows there are finitely many of
the sets V_{h}, V_{h1}
⋅⋅⋅
V_{hn} such that H ⊆∪_{i=1}^{n}V_{hi}. Then letting U = ∩_{i=1}^{n}U_{hi} and
V = ∪_{i=1}^{n}V_{hi}, it follows that p ∈ U, H ∈ V and U ∩ V = ∅. Thus X is regular as
claimed.
Next let K and H be disjoint nonempty closed sets.Using regularity of X, for every
k ∈ K, there exists an open set U_{k} containing k and an open set V_{k} containing H such
that these two open sets have empty intersection. Thus H ∩U_{k} = ∅. Finitely many of the
U_{k}, U_{k1},
⋅⋅⋅
,U_{kp} cover K and so ∪_{i=1}^{p}U_{ki} is a closed set which has empty
intersection with H. Therefore, K ⊆∪_{i=1}^{p}U_{ki} and H ⊆
(∪p U- )
i=1 ki
^{C}. This proves the
theorem.
A useful construction when dealing with locally compact Hausdorff spaces is the
notion of the one point compactification of the space discussed earler. However, it is
reviewed here for the sake of convenience or in case you have not read the earlier
treatment.
Definition 10.2.5Suppose
(X,τ)
is a locally compact Hausdorff space. Then let
X^
≡ X ∪
{∞}
where ∞ is just the name of some point which is not in X which is calledthe point at infinity. A basis for the topology
^τ
for
^X
is
{ C }
τ ∪ K where K is a compact subset of X .
The complement is taken with respect to
X^
and so the open sets, K^{C}are basic open setswhich contain ∞.
The reason this is called a compactification is contained in the next lemma.
Lemma 10.2.6If
(X,τ)
is a locally compact Hausdorff space, then
( ^ )
X,^τ
is acompact Hausdorff space. Also if U is an open set of
^τ
, then U ∖
{∞ }
is an openset of τ.
Proof:Since
(X, τ)
is a locally compact Hausdorff space, it follows
( )
X^, ^τ
is a
Hausdorff topological space. The only case which needs checking is the one of p ∈ X and
∞. Since
(X, τ)
is locally compact, there exists an open set of τ, U having compact
closure which contains p. Then p ∈ U and ∞∈U^{C} and these are disjoint open sets
containing the points, p and ∞ respectively. Now let C be an open cover of
X^
with sets from
^τ
. Then ∞ must be in some set, U_{∞} from C, which must
contain a set of the form K^{C} where K is a compact subset of X. Then there exist
sets from C, U_{1},
⋅⋅⋅
,U_{r} which cover K. Therefore, a finite subcover of
^X
is
U_{1},
⋅⋅⋅
,U_{r},U_{∞}.
To see the last claim, suppose U contains ∞ since otherwise there is nothing to
show. Notice that if C is a compact set, then X ∖ C is an open set. Therefore, if
x ∈ U ∖
{∞ }
, and if
^X
∖ C is a basic open set contained in U containing ∞, then if x
is in this basic open set of
^X
, it is also in the open set X ∖ C ⊆ U ∖
{∞}
.
If x is not in any basic open set of the form
^X
∖ C then x is contained in an
open set of τ which is contained in U ∖
{∞ }
. Thus U ∖
{∞ }
is indeed open in
τ.
Theorem 10.2.7Let X be a locally compact Hausdorff space, and let K bea compact subset of the open set V . Then there exists a continuous function,f : X →
[0,1]
, such that f equals 1 on K and
{x : f (x) ⁄= 0}
≡spt
(f)
is a compactsubset of V .
Proof: Let
^
X
be the space just described. Then K and V are respectively closed and
open in
,U, and W such that
K ⊆ U,∞∈ V^{C}⊆ W, and U ∩ W = U ∩
(W ∖{∞ })
= ∅.
PICT
Thus W ∖
{∞ }
is an open set in the original topological space which contains V^{C},U
is an open set in the original topological space which contains K, and W ∖
{∞ }
and U
are disjoint.
Now for each x ∈ K, let U_{x} be a basic open set whose closure is compact and such
that
x ∈ U ⊆ U.
x
Thus U_{x} must have empty intersection with V^{C} because the open set, W ∖
{∞ }
contains no points of U_{x}. Since K is compact, there are finitely many of these sets,
U_{x1},U_{x2},
⋅⋅⋅
,U_{xn} which cover K. Now let H ≡∪_{i=1}^{n}U_{xi}.
Claim: H = ∪_{i=1}^{n}U_{xi}
Proof of claim: Suppose p ∈H. If p
∕∈
∪_{i=1}^{n}U_{xi} then if follows p
∈∕
U_{xi} for each
i. Therefore, there exists an open set, R_{i} containing p such that R_{i} contains
no other points of U_{xi}. Therefore, R ≡∩_{i=1}^{n}R_{i} is an open set containing p
which contains no other points of ∪_{i=1}^{n}U_{xi} = W, a contradiction. Therefore,
H⊆∪_{i=1}^{n}U_{xi}. On the other hand, if p ∈U_{xi} then p is obviously in H so this proves the
claim.
From the claim, K ⊆ H ⊆H⊆ V and H is compact because it is the finite union of
compact sets. By Urysohn’s lemma, there exists f_{1} continuous on H which has values in
[0,1]
such that f_{1} equals 1 on K and equals 0 off H. Let f denote the function which
extends f_{1} to be 0 off H. Then for α > 0, the continuity of f_{1} implies there exists U open
in the topological space such that
− 1 −1 --C ( --) -C -C
f ((− ∞,α )) = f1 ((− ∞, α))∪ H = U ∩ H ∪ H = U ∪ H
an open set. If α ≤ 0,
f −1((− ∞, α)) = ∅
an open set. If α > 0, there exists an open set U such that
−1 −1 --
f ((α,∞ )) = f1 ((α,∞ )) = U ∩ H = U ∩ H
because U must be a subset of H since by definition f = 0 off H. If α ≤ 0,
then
−1
f ((α, ∞)) = X,
an open set. Thus f is continuous and spt
(f )
⊆H, a compact subset of V. This proves
the theorem.
In fact, the conclusion of the above theorem could be used to prove that the
topological space is locally compact. However, this is not needed here.
In case you would like a more elementary proof which does not use the one point
compactification idea, here is such a proof.
Theorem 10.2.8Let X be a locally compact Hausdorff space, and let K bea compact subset of the open set V . Then there exists a continuous function,f : X →
[0,1]
, such that f equals 1 on K and
{x : f (x) ⁄= 0}
≡spt
(f)
is a compactsubset of V .
Proof: To begin with, here is a claim. This claim is obvious in the case of a metric
space but requires some proof in this more general case.
Claim:If k ∈ K then there exists an open set U_{k} containing k such that U_{k} is
contained in V.
Proof of claim: Since X is locally compact, there exists a basis of open sets whose
closures are compact, U. Denote by C the set of all U ∈U which contain k
and let C^{′} denote the set of all closures of these sets of C intersected with the
closed set V^{C}. Thus C^{′} is a collection of compact sets. I will argue that there
are finitely many of the sets of C^{′} which have empty intersection. If not, then
C^{′} has the finite intersection property and so there exists a point p in all of
them. Since X is a Hausdorff space, there exist disjoint basic open sets from
U, A,B such that k ∈ A and p ∈ B. Therefore, p
∕∈
A contrary to the above
requirement that p be in all such sets. It follows there are sets A_{1},
⋅⋅⋅
,A_{m} in C such
that
C --- ---
V ∩ A1 ∩ ⋅⋅⋅∩ Am = ∅
Let U_{k} = A_{1}∩
⋅⋅⋅
∩A_{m}. Then U_{k}⊆A_{1}∩
⋅⋅⋅
∩A_{m} and so it has empty intersection with
V^{C}. Thus it is contained in V . Also U_{k} is a closed subset of the compact set A_{1} so it is
compact. This proves the claim.
Now to complete the proof of the theorem, since K is compact, there are finitely
many U_{k} of the sort just described which cover K,U_{k1},
⋅⋅⋅
,U_{kr}. Let
r
H = ∪i=1Uki
so it follows
-- ---
H = ∪ri=1Uki
and so K ⊆ H ⊆H⊆ V and H is a compact set. By Urysohn’s lemma, there exists f_{1}
continuous on H which has values in
[0,1]
such that f_{1} equals 1 on K and equals 0 off
H. Let f denote the function which extends f_{1} to be 0 off H. Then for α > 0,
the continuity of f_{1} implies there exists U open in the topological space such
that
-- -- -- --
f− 1((− ∞,α )) = f−11((− ∞, α))∪ HC = (U ∩ H )∪ HC = U ∪ HC
an open set. If α ≤ 0,
f −1((− ∞, α)) = ∅
an open set. If α > 0, there exists an open set U such that
f−1((α,∞ )) = f−1((α,∞ )) = U ∩ H-= U ∩ H
1
because U must be a subset of H since by definition f = 0 off H. If α ≤ 0,
then
f −1((α, ∞)) = X,
an open set. Thus f is continuous and spt
(f )
⊆H, a compact subset of V. This proves
the theorem.
Definition 10.2.9Define spt(f) (support of f)to be the closure of the set{x : f(x)≠0}. If V is an open set, C_{c}(V ) will be the set of continuous functions f,defined on Ω having spt(f) ⊆ V . Thus in Theorem 10.2.7or 10.2.8, f ∈ C_{c}(V ).
Definition 10.2.10If K is a compact subset of an open set, V , then K ≺ ϕ ≺ Vif
ϕ ∈ Cc(V),ϕ(K ) = {1},ϕ(Ω) ⊆ [0,1],
where Ω denotes the whole topological space considered. Also for ϕ ∈ C_{c}(Ω), K ≺ ϕif
ϕ(Ω) ⊆ [0,1] and ϕ (K ) = 1.
and ϕ ≺ V if
ϕ(Ω) ⊆ [0,1] and spt(ϕ) ⊆ V.
Theorem 10.2.11(Partition of unity)Let K be a compact subset of a locallycompact Hausdorff topological space satisfying Theorem 10.2.7or 10.2.8andsuppose
n
K ⊆ V = ∪i=1Vi,Vi open.
Then there exist ψ_{i}≺ V_{i}with
∑n
ψi(x ) = 1
i=1
for all x ∈ K.
Proof: Let K_{1} = K ∖∪_{i=2}^{n}V_{i}. Thus K_{1} is compact and K_{1}⊆ V_{1}. Let
K_{1}⊆ W_{1}⊆W_{1}⊆ V_{1}with W_{1}compact. To obtain W_{1}, use Theorem 10.2.7 or 10.2.8 to
get f such that K_{1}≺ f ≺ V_{1} and let W_{1}≡
{x : f (x) ⁄= 0}
.Thus W_{1},V_{2},
⋅⋅⋅
V_{n} covers
K and W_{1}⊆ V_{1}. Let K_{2} = K ∖ (∪_{i=3}^{n}V_{i}∪W_{1}). Then K_{2} is compact and K_{2}⊆ V_{2}. Let
K_{2}⊆ W_{2}⊆W_{2}⊆ V_{2}W_{2} compact. Continue this way finally obtaining W_{1},
⋅⋅⋅
,W_{n},
K ⊆ W_{1}∪
⋅⋅⋅
∪ W_{n}, and W_{i}⊆ V_{i}W_{i} compact. Now let W_{i}⊆ U_{i}⊆U_{i}⊆ V_{i},U_{i}
compact.
PICT
By Theorem 10.2.7 or 10.2.8, let U_{i}≺ ϕ_{i}≺ V_{i},∪_{i=1}^{n}W_{i}≺ γ ≺∪_{i=1}^{n}U_{i}.
Define
∪_{i=1}^{n}U_{i}. Consequently γ(y) = 0 for all y
near x and so ψ_{i}(y) = 0 for all y near x. Hence ψ_{i} is continuous at such x. If
∑_{j=1}^{n}ϕ_{j}(x)≠0, this situation persists near x and so ψ_{i} is continuous at such points.
Therefore ψ_{i} is continuous. If x ∈ K, then γ(x) = 1 and so ∑_{j=1}^{n}ψ_{j}(x) = 1. Clearly
0 ≤ ψ_{i}
(x)
≤ 1 and spt(ψ_{j}) ⊆ V_{j}. This proves the theorem.
The following corollary won’t be needed immediately but is of considerable interest
later.
Corollary 10.2.12If H is a compact subset of V_{i}, there exists a partition of unitysuch that ψ_{i}
(x)
= 1 for all x ∈ H in addition to the conclusion of Theorem 10.2.11.
Proof:Keep V_{i} the same but replace V_{j} with
^Vj
≡ V_{j}∖H. Now in the proof above,
applied to this modified collection of open sets, if j≠i,ϕ_{j}