General Topology — Part 7 (Locally Compact Hausdroff Space)

6 min readNov 27, 2023

In the Locally Compact Hausdroff Space, when a compact set is contained by an open set, a “bump” function over this open set will exist. To prove this theorem, we need the following definition and lemma first :

X denotes the topological space
f:X→R denotes the real-valued continuous function on X
C(X) denotes space of f on X
supp denotes the Support of f where f ∈ C(x):
supp(f) = Cl({ x ∈ X| f(x)≠0 })

The Support of a function is defined as the closure of the set of points where the function is non-zero. The closure is required here since this can ensure that isolated points where the function is non-zero are also included. Remember that the closure of a set includes all the limit point and the given set.

K(X) denotes the set of f ∈ C(x) having Compact Support:
K(X) = { f ∈ C(x) | supp(f) is compact }

I denotes the indicator function:
Given a subset A ⊆ X, IA: X → {0, 1}
IA(x) = 1 if x ∈ A
IA(x) = 0 if x ∉ A

Lemma A ( from Part 4 )
Let X be a normal space. For any closed set A ⊆ X and an open set C ∈ τ where A ⊆ C, there is an open set o∈τ s.t. A ⊆ o ⊆ Cl(o) ⊆ C

LCH1. Let X be a Locally Compact Hausdorff space. For any compact set k⊂X and let U be open set in X where k ⊆ U, there exists f ∈ K(X), satisfies Ik≤ f ≤ IU and k ⊆ supp(f) ⊆ U

Proof-
U is open in X
⇒ U is open in X*
[ X* is the one-point compactification extension of X, so τ ⊂ τ* and X* is Compact Hausdorff ]
k is compact in X
⇒ k is compact in X* [ By CP4 where X is subspace of X* ]
X* is Hausdorff and k is compact in X* [ Since X* is Compact Hausdorff ]
⇒ k is closed in X* [ By HD1 ]
X* is Compact Hausdorff ⇒ X* is Normal [ By CH1 ]
⇒ ∃ o∈τ* s.t. k ⊆ o ⊆ ClX*(o) ⊆ U [ By Lemma A ]

k is closed in X*, X*\o is closed in X* and X* is Normal
⇒ ∃ g: X* → [0, 1] is continuous where g(k)=1, g(X*\o)=0 [ By Urysohn’s lemma in Part 4 ]
Let f: X → [0, 1], f = g|X is continuous [ Since Restriction function is continuous ]
f(k)=g(k)=1, f(X\o)=g(X\o)=0 [ Since X\o ⊂ X*\o ]
If x ∈ k ⇒ f(x)=1, Ik(x)=1, IU(x)=1 ⇒ Ik = f = IU
If x ∈ X\o ⇒ f(x)=0, Ik(x)=0, IU(x)=0 or 1 ⇒ Ik ≤ f ≤ IU
If x ∈ o\k ⇒ 0<f(x)<1, Ik(x)=0, IU(x)=1 ⇒ Ik < f < IU
⇒ Ik ≤ f ≤ IU
supp(f) = ClX({x ∈ X| f(x)≠0}) = ClX({x ∈ X| x∈ k ∪ (o\k)}) = ClX(o) [ Since k ⊆ o]
ClX*(o) is closed in X* [ By CL2, closure is smallest closed set containing o ]
⇒ ClX*(o) is compact in X* [ By CP1 where X* is compact ]
⇒ ClX*(o) is compact in X [ By CP4, compactness is preserved ]
ClX*(o) = ClX(o) [ By CL12, ClX(o) = X∩CIX*(o) = CIX*(o) since CIX*(o)⊆U⊆X ]
⇒ ClX(o) is compact in X
⇒ k ⊆ supp(f) = ClX(o) = ClX*(o) ⊆ U
⇒ supp(f) is compact ⇒ f has compact support

If the support of a continuous function is compact and can be contained by a finite union of open sets, then we can decompose this function as the sum of bump functions with compact support which is contained by one of the open sets.

LCH2. Let X be a Locally Compact Hausdorff space. Let f ∈ K(X) and let U1,U2, ... Un be open subsets of X s.t. supp(f) ⊆ ∪{ Ui | 1≤i≤n }. Then, there are continuous f1,..,fn ∈ K(X) s.t. f = f1+..+fn and for each i, supp(fi) ⊆ Ui. Furthermore, if f is non-negative, then each fi can be chosen to be non-negative as well.

Proof-
Let us prove the LCH2 statement is true by induction.
When n = 1, the statement is true.

When n = 2:
supp(f) ⊆ U1∪U2
⇒ There exists compact sets K1, K2 s.t. supp(f)=K1∪K2 where K1⊆U1, K2⊆U2 [ By HD5 where X is Hausdorff, f ∈ K(X) and supp(f) is compact ]
⇒ There exists g1,g2 ∈ K(X), satisfies IK1≤ g1 ≤ IU1, IK2≤ g2 ≤ IU2 and K1⊆supp(g1)⊆ U1, K2⊆supp(g2)⊆ U2 [ By LCH1 ]

Let f1:X→R and f2:X->R be the real-valued functions on X: f1(x)=g1(x)/(g1(x)+g2(x))*f(x) if x ∈ supp(g1)
f1(x)=0 otherwise
f2(x)=g2(x)/(g1(x)+g2(x))*f(x) if x ∈ supp(g2)
f2(x)=0 otherwise [ The set {f1,f2} is also called Partition of Unity ]
⇒
If x ∈ supp(g1), g1/g1+g2 is the elementary combination of continuous function
If x ∉ supp(g1), f1=0 is a constant function which is continuous [ see part 2 ]
⇒
f1 is continuous in both cases
f2 is continuous [ By similar argument ] and f1,f2 are non-negative when f is non-negative.

supp(f1) = Cl({ x ∈ X| f1(x)≠0 }) = Cl({x ∈ X| f≠0} ∩ {x ∈ X| g1≠0})
⇒ supp(f1) ⊆ Cl({x ∈ X| f≠0}) ∩ Cl({x ∈ X| g1≠0}) = supp(f)∩supp(g1) [ By CL10, note that g1,g2 ≥0, g1+g2 ≠0 when g1≠0 ]
⇒ supp(f1) ⊆ supp(g1) ⊆ U1 and supp(f1) is compact [ Since supp(f1) is closed and supp(g1) is compact and by CP1 ]
⇒ supp(f2) ⊆ supp(g2) ⊆ U2 and supp(f2) is compact [ By similar argument ]
⇒ f1, f2 ∈ K(X)

If x ∉ supp(f)
⇒ f = 0
⇒ if x∈supp(g1), f1=g1/(g1+g2)*0, if x∉supp(g1), f1=0 ⇒ f1=0 in both cases
f2 = 0 [ By similar argument ]
⇒ f1+f2 = f = 0

If x ∈ supp(f)
If x ∈ supp(g1)∩supp(g2)
⇒ f1+f2 = [g1/(g1+g2) + g2/(g1+g2)]*f= f
If x ∈ supp(g1) ∩ X\supp(g2)
⇒ f1+f2 = g1/(g1+0)*f + 0 = f
If x ∈ X\supp(g1) ∩ supp(g2)
⇒ f1+f2 = 0 + g2/(0+g2)*f = f
If x ∈ X\supp(g1) ∩ X\supp(g2)
⇒ f1+f2 = 0 + 0 = 0, x ∈ X\K1∩X\K2=X\(K1∪K2)=X\supp(f)
[ since K1⊆supp(g1) and K2⊆supp(g2) ]
⇒ f = 0 , f1+f2=0 ⇒ f1+f2=f
⇒ f = f1+f2 in all cases
Hence, the statement is true for n=2

Assume the statement is true for n=k.

When n = k+1, there are U1,U2, … Uk+1 open subsets of X s.t. supp(f) ⊆ ∪{Ui | 1≤i≤k+1 }.

Let U=∪{ Ui | 1≤i≤k }
⇒ U, Uk+1 are open sets and supp(f) ⊆ U∪Uk+1 [ By T2, Part 1 ]
⇒ There exist continuous h,fk+1 ∈ K(X) s.t. f = h + fk+1, supp(h) ⊆ U and supp(fk+1) ⊆ Uk+1 and h,fk+1 are non-negative when f is non-negative [Since statement is true when n=2 ]

h ∈ K(X) and U1,U2, … Uk are open subsets of X and supp(h) ⊆∪{ Ui |1≤i≤k}
⇒ There exist continuous f1,..,fk ∈ K(X) s.t. h = f1+..+fk and for each i where 1≤i≤k, supp(fi) ⊆ Ui and f1,..fk are non-negative as h is non-negative [ Since statement is true when n=k ]

⇒ There exist continuous f1,..,fk,fk+1 ∈ K(X) s.t. f = f1+..+fk+fk+1 and for each i where 1≤i≤k+1, supp(fi) ⊆ Ui and f1,..fk,fk+1 are non-negative when f is non-negative
⇒ The statement is also true for n=k+1
⇒ Hence, by induction, the statement is true for n=1,2,..

Conclusion

In this post, we have found 2 nice properties of the Locally Compact Hausdroff Space. 1) When a compact set in this space is contained by an open set, a “bump” function over this open set will exist. Outside this open set, the function will become zero. 2) If the support of a continuous function is compact and can be contained by a finite union of open sets, then we can decompose this function as the sum of bump functions with compact support.

In the next article, I will list all the theorems we have learned so far.