11. More on locally compact Hausdorff spaces - Aarhus Universitet

<strong>11.</strong> <strong>More</strong> on locally compact Hausdorff spaces 

Klaus Thomsen matkt@imf.au.dk 

Institut for Matematiske Fag 

Det Naturvidenskabelige Fakultet 

Aarhus Universitet 

November 2005 

Klaus Thomsen <strong>11.</strong> <strong>More</strong> on locally compact Hausdorff spaces

Additional material, comes from ’Noter og kommentarer’ til Rudins 

bog. 


The support of a continuous function 

Let X be a locally compact Hausdorff space. The support of a 

function f : X → C is the set 

supp f = {x ∈ X : f (x) = 0}; 

i.e. the closure of the set of points where f is not zero. 

We say that f : X → C has compact support when supp f is 

compact. 

The following famous lemma, Urysohn’s lemma, guarantees a rich 

supply of continuous functions of compact support on a locally 

compact Hausdorff space. 


Urysohn’s lemma - the statement 

Lemma 

(Urysohn’s lemma) Let X be a locally compact Hausdorff space, K 

a compact subset of X and V an open subset of X such that 

K ⊆ V . It follows that there is a continuous function f : X → [0, 1] 

of compact support such that f (k) = 1, k ∈ K, and supp f ⊆ V . 

If d is a metric for the topology (which in the generality of the 

lemma may not exist), one can construct f relatively painlessly: For 

x ∈ X , set 

Dist(x, K) = inf {d(x, y) : y ∈ K} . 

Choose (!) ɛ > 0 so small that {x ∈ X : Dist(x, K) ≤ ɛ} ⊆ V , and 

set 

f (y) = max 1 − ɛ −1 Dist(y, K), 0 

for all y ∈ X . 

It is not difficult to see that this function has the stated properties, 

at least when we assume, as we may, that V has compact closure. 

The general (non-metric) case is a little more difficult. 


Urysohn’s lemma - the proof 

Let r1, r2, r3, . . . be a list of the rational numbers in ]0, 1[, chosen 

such that r1 = 0 and r2 = 1, and without repetitions. 

It follows from a lemma from Additional material 1 that there are 

open sets V0 and V1 in X with V0 compact such that 

K ⊆ V1 ⊆ V1 ⊆ V0 ⊆ V0 ⊆ V . 

Note that V1 is then also compact. Suppose now that n ≥ 2, and 

that we have found open sets Vr1 , Vr2 , Vr3 , . . . , Vrn in X such that 

Vri is compact for all i, and 

ri < rj ⇒ K ⊆ Vrj 

⊆ Vrj ⊆ Vri ⊆ Vri ⊆ V . 

Let ri be the largest element of {r1, r2, . . . , rn} which is smaller 

than rn+1, and rj the smallest element of {r1, r2, . . . , rn} which is 

larger than rn+1. 



We can then use the same lemma as above to find an open set 

in X such that 

Vrn+1 

Vrj 

⊆ Vrn+1 ⊆ Vrn+1 ⊆ Vri . 

Continuing in this way we obtain a collection {Vr } of open sets in 

X , one for each rational number r ∈ (0, 1) such that Vr is 

compact, K ⊆ V1, V0 ⊆ V , and 

r < s ⇒ Vs ⊆ Vr . (1) 

When r, s ∈]0, 1[ are rationals, define fr : X → [0, 1] and 

gs : X → [0, 1] such that 

 

fr (x) = 

r, 

0, 

if x ∈ Vr 

otherwise 

and 

gs(x) = 

 

1, if x ∈ Vs 

s, otherwise 



Set 

and 

f (x) = sup fr (x) 

r 

g(x) = inf 

s gs(x). 

If x ∈ X is a point where fr (x) > gs(x), it follows from the 

definitions that x ∈ Vr and x /∈ Vs, 

and hence from (1) that s > r. 

But this is impossible since gs(x) ≥ s and fr (x) ≤ r. 

We see therefore that fr (x) ≤ gs(x) for all r, s, and hence that 

f (x) ≤ g(x) for all x ∈ X . 

If x ∈ X is a point such that f (x) < g(x), there are rational 

numbers r, s ∈]0, 1[ such that f (x) < r < s < g(x). 

The first inequality implies that x /∈ Vr and the second that x ∈ Vs. 

This impossible by (1), and we conclude therefore that f = g. 



Since K ⊆ Vr for all r, we see that f (k) = 1 when k ∈ K and since 

fr (x) = 0 for all r when x /∈ V0, we see that supp f ⊆ V0 so that f 

has compact support. 

It remains now only to show that f is continuous. To this end it 

suffices to show that f −1 (]α, β[) is open when α < β. Note first 

that 

and that 

g −1 (] − ∞, β[) = 

g −1 

r (] − ∞, β[) = 

This shows that g −1 (] − ∞, β[) is open. 

r 

g −1 

r (] − ∞, β[), 

⎧ 

⎪⎨ X , when β > 1 

c 

Vr , when r < β ≤ 1, . 

⎪⎩ 

∅, when β ≤ r 



Similarly, 

where 

f −1 (]α, ∞[) = 

f −1 

r (]α, ∞[) = 

r 

f −1 

r (]α, ∞[), 

⎧ 

⎪⎨ ∅, when α ≥ r, 

Vr , 

⎪⎩ 

X , 

when 0 ≤ α < r, 

when α < 0. 

This shows that f −1 (]α, ∞[) is open. Since 

f −1 (]α, β[) = g −1 (] − ∞, β[) ∩ f −1 (]α, ∞[), 

we see that f −1 (]α, β[) is open, as desired. 


Partition of unity - the statement 

Theorem 

Let X be locally compact Hausdorff space, K a compact subset of 

X , and V1, V2, . . . , Vn open subsets of X such that 

K ⊆ V1 ∪ V2 ∪ · · · ∪ Vn. 

It follows that there are continuous functions 

hi : X → [0, 1], i = 1, 2, . . . , n, all of compact supports such that 

supp hi ⊆ Vi, for all i, and 

for all k ∈ K. 

n 

hi(k) = 1 

i=1 

This is often expressed by saying that ’{hi} is a partition of unity 

on K subordinate to the cover {Vi}’. 


Partition of unity - the proof 

Let x ∈ K. It follows from the lemma from Additional material 1 

that there is an i ∈ {1, 2, . . . , n}, and an open neighborhood Wx of 

x such that Wx is compact and 

x ∈ Wx ⊆ Wx ⊆ Vi. 

By compactness of K there are then points x1, x2, . . . , xm in K such 

that K ⊆ Wx1 ∪ Wx2 ∪ · · · ∪ Wxm. Let Hj be the union of the Wx k ’s 

that are subsets of Vj. 

It follows from Urysohn’s lemma that there are continuous 

functions gj : X → [0, 1], j = 1, 2, . . . , n, of compact supports such 

that gj(y) = 1, y ∈ Hj and supp gj ⊆ Vj for all j. 


Partition of unity - the proof 

Set 

h1 = g1, 

h2 = (1 − g1) g2, 

h3 = (1 − g1) (1 − g2) g3, 

. 

hn = (1 − g1) (1 − g2) . . . (1 − gn−1) gn. 

Then supp hi ⊆ supp gi is compact and a subset of Vi for all i. 

It is easy to prove, e.g. by induction, that 

h1 + h2 + · · · + hn = 1 − (1 − g1)(1 − g2)(1 − g3) . . . (1 − gn) 

so we see that n 

i=1 hi(k) = 1 when k ∈ K. 


The Stone-Weierstrass theorem - the classical theorem 

We prove here a version of the Stone-Weierstrass theorem. The 

point of departure is the classical theorem of Weierstrass: 

Theorem 

Let [a, b] be a compact interval in the real line R. Every continuous 

function on [a, b] can approximated uniformly by polynomials, i.e. 

for every f ∈ C[a, b] and every ɛ > 0 there is a polynomial P such 

that 

sup |f (t) − P(t)| < ɛ. 

t∈[a,b] 

We assume that the reader is familiar with this classical theorem. 

We are only going to use it in the rather special case where 

f (t) = |t| on [−b, b], for some b > 0, so if the reader knows 

another way of approximating this function with polynomials, the 

following will give a self-contained proof of Weierstrass’ theorem. 


The Stone-Weierstrass theorem - the general case 

Theorem 

Let X be a compact Hausdorff space. Let A ⊆ C(X ) be a 

subalgebra of C(X ) such that 

a) A is self-adjoint, i.e. f ∈ A ⇒ f ∈ A, 

b) A separates points on X , i.e. when x, y ∈ X and x = y, then 

there is an element g ∈ A such that g(x) = g(y), 

c) A vanishes at no point, i.e. for every x ∈ X there is an f ∈ A 

such that f (x) = 0. 

It follows that A is dense in C(X ), i.e. for all g ∈ C(X ) and ɛ > 0 

there is an element f ∈ A such that 

sup |g(x) − f (x)| < ɛ. 

x∈X 


The Stone-Weierstrass theorem - the proof in general case 

Set AR = {f ∈ A : f (x) ∈ R ∀x ∈ X }. For every g ∈ C(X ), 

g = Re g + i Im g, where 

Re g = 1 

(g + g) 

2 

and 

Im g = 1 

(g − g) . 

2i 

It follows from a) that A = AR + iAR. It suffices therefore to show 

that a real-valued function f ∈ C(X ) can be approximated 

uniformly by elements of AR. 

To this end, note that AR separates points and does not not vanish 

at any point, since this is true for A by assumption. 



We prove first that 

∀ x1, x2 ∈ X , x1 = x2, ∀ r1, r2 ∈ R ∃g ∈ AR : g(xi) = ri, i = 1, 2. 

(2) 

Indeed, it follows from b) that there is a g ∈ AR such that 

g(x1) = g(x2). It follows from c) that there are functions 

h, k ∈ AR such that h(x1) = 0 and k(x2) = 0. Put 

u = gk − g(x1)k, v = gh − g(x2)h, and note that u, v ∈ AR. 

Since u(x1) = v(x2) = 0, while u(x2) = 0 and v(x1) = 0, we can 

set 

g = r1v r2u 

+ 

v(x1) u(x2) . 

Then g ∈ AR and g(xi) = ri, i = 1, 2, proving (2). 



Next we prove that 

f ∈ AR ⇒ |f | ∈ AR. (3) 

(Recall that AR is the closure of AR in C(X ).) 

To prove (3), it suffices to take an ɛ > 0 and find h ∈ AR such that 

sup x∈X |h(x) − |f (x)|| < 2ɛ. 

To this end note that since f ∈ AR, there is an f1 ∈ AR such that 

sup ||f (x)| − |f1(x)|| ≤ sup |f (x) − f1(x)| < ɛ. (4) 

x∈X 

x∈X 

By Theorem 3 there is a polynomial P such that 

|P(t) − |t|| < ɛ (5) 

for all t ∈ [−f1, f1]. 

Using 1 

 

2 P + P in place of P, we may assume that P is a 

real-valued polynomial. It follows then that h = P ◦ f1 is in AR 

because A is a subalgebra of C(X ). 



It follows from (5) that sup x∈X |h(x) − |f1(x)|| < ɛ. By combining 

this with (4), we conclude that sup x∈X |h(x) − |f (x)|| < 2ɛ, 

completing the proof of (3). 

It follows from (3) that 

because max{f , g} = 

f , g ∈ AR ⇒ max{f , g}, min{f , g} ∈ AR, (6) 

f +g 

2 

+ |f −g| 

2 

and min{f , g} = 

f +g 

2 

− |f −g| 

2 . 



Next we prove that 

I) for any real-valued f ∈ C(X ), any x ∈ X and any ɛ > 0, there 

is a function gx ∈ AR such that gx(x) = f (x) and 

gx(y) > f (y) − ɛ for all y ∈ X . 

To this end, note that it follows from (2) that for every y ∈ X , 

there is a hy ∈ AR such that hy (x) = f (x) and hy (y) = f (y). 

There is then an open neighborhood Jy of y such that 

hy (z) > f (z) − ɛ 

for all z ∈ Jy . 

Since X is compact there are finitely many Jyi , i = 1, 2, . . . , N, such 

that X ⊆ Jy1 ∪ Jy2 ∪ · · · ∪ JyN . 

It follows from (6) that 

gx = max {hy1 , hy2 , . . . , hyN } 

is in AR. 

Furthermore, by construction gx(x) = f (x) and gx(y) > f (y) − ɛ 

for all y ∈ X , completing the proof of I). 



The proof of the theorem is then completed by using I) in the 

following way: When f ∈ C(X ) is real-valeud and ɛ > 0, it follows 

from I) that for every x ∈ X there is a function gx ∈ AR such that 

gx(x) = f (x) and gx(y) > f (y) − ɛ for all y ∈ X . 

For each x ∈ X there is then an open neighborhood Ix of X such 

that 

gx(y) < f (y) + ɛ 

for all y ∈ Ix. 

Since X is compact there are finitely many x1, x2, . . . , xM ∈ X such 

that X ⊆ Ix1 ∪ Ix2 ∪ · · · ∪ IxM . 

Set 

g = min {gx1 , gx2 , . . . , gxM } , 

and note that g ∈ AR. Since 

f (z) − ɛ < g(z) < f (z) + ɛ 

for all z ∈ X , this completes the proof.

11. More on locally compact Hausdorff spaces - Aarhus Universitet

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