§ 4. Differential Manifolds
267
and in setting by definition,
X
ω =
Ui
ω i
(17.3)
since, in any reasonable interpretation of the integral of ω i , it suffices to
integrate over the open subset U i outside which ω i vanishes. But to give an
“ absolute ” meaning to the left hand side of (3), the right hand side still
needs to be shown to remain invariant if the U i are replaced by the open
Cartesian subsets V p covering K and the ω i by the ω p satisfying (2) for the
new covering, which is not at all obvious. Partitions of unity need to be used
for this, a technique that can prove useful elsewhere.
Lemma 1. Let A and B be two disjoint closed subsets in a metric space X.
There is a function f defined and continuous on X satisfying f (x) = 1 on
A, f (x) = 2 on B and 1 ≤ f (x) ≤ 2 everywhere.
The lemma is trivial if A is empty (take f = 2 everywhere) or if B is
empty (take f = 1 everywhere). Otherwise, choose a distance function d(x, y)
defining the topology on X and set
f (x) = inf [d(x, A), 2d(x, B)] / inf [d(x, A), d(x, B)]
for x ∈ X − (A ∪ B), an open set on which f is continuous since so are the
functions d(x, A) and d(x, B) that do not vanish there. In the neighbourhood
of every point of A, d(x, A) < d(x, B) and so f (x) = 1 ; in the neighbourhood
of every point of B, 2d(x, B) < d(x, A) and so f (x) = 2. Setting f (x) = 1 on
A and f (x) = 2 on B, we find the function sought
68 in all of the space X.
Lemma 2. Let U be an open set and A a closed one contained in U . There
is an open set V such that A ⊂ V ⊂ ¯
V ⊂ U . If X is locally compact
69 and A
is compact, ¯
V may be assumed to be compact.
Here too, the first statement is trivial if U = X (take V = X) or if A
is empty (take V = {x} with x ∈ U ). Otherwise, set B = X − U , choose a
function f by lemma 1 and take V = {f (x) < 3/2}, an open set containing
A trivially and whose closure, contained in the open set {f (x) ≤ 3/2}, does
not meet B = X − U ; so ¯
V ⊂ U .
68 A more general result (Urysohn’s theorem): if f is a real, bounded continuous
function defined on a closed set F ⊂ X, there is a continuous extension of f to
X. Assuming f (F ) ⊂ [1, 2], the case it reduces to, formula
f (x) = d(x, F )
−1 . inf [f (u)d(x, u)] for x ∈ X − F ,
where inf relates to the u ∈ F , provides a solution. Dieudonn´ e, IV, 5. The lemma
correspond to the case F = A ∪ B.
69 i.e. such that every x ∈ X has a compact neighbourhood V . Then any neighbourhood of x contains a compact neighbourhood, for example V ∩ B, where B
is a closed ball centered at x contained in W .
267
and in setting by definition,
X
ω =
Ui
ω i
(17.3)
since, in any reasonable interpretation of the integral of ω i , it suffices to
integrate over the open subset U i outside which ω i vanishes. But to give an
“ absolute ” meaning to the left hand side of (3), the right hand side still
needs to be shown to remain invariant if the U i are replaced by the open
Cartesian subsets V p covering K and the ω i by the ω p satisfying (2) for the
new covering, which is not at all obvious. Partitions of unity need to be used
for this, a technique that can prove useful elsewhere.
Lemma 1. Let A and B be two disjoint closed subsets in a metric space X.
There is a function f defined and continuous on X satisfying f (x) = 1 on
A, f (x) = 2 on B and 1 ≤ f (x) ≤ 2 everywhere.
The lemma is trivial if A is empty (take f = 2 everywhere) or if B is
empty (take f = 1 everywhere). Otherwise, choose a distance function d(x, y)
defining the topology on X and set
f (x) = inf [d(x, A), 2d(x, B)] / inf [d(x, A), d(x, B)]
for x ∈ X − (A ∪ B), an open set on which f is continuous since so are the
functions d(x, A) and d(x, B) that do not vanish there. In the neighbourhood
of every point of A, d(x, A) < d(x, B) and so f (x) = 1 ; in the neighbourhood
of every point of B, 2d(x, B) < d(x, A) and so f (x) = 2. Setting f (x) = 1 on
A and f (x) = 2 on B, we find the function sought
68 in all of the space X.
Lemma 2. Let U be an open set and A a closed one contained in U . There
is an open set V such that A ⊂ V ⊂ ¯
V ⊂ U . If X is locally compact
69 and A
is compact, ¯
V may be assumed to be compact.
Here too, the first statement is trivial if U = X (take V = X) or if A
is empty (take V = {x} with x ∈ U ). Otherwise, set B = X − U , choose a
function f by lemma 1 and take V = {f (x) < 3/2}, an open set containing
A trivially and whose closure, contained in the open set {f (x) ≤ 3/2}, does
not meet B = X − U ; so ¯
V ⊂ U .
68 A more general result (Urysohn’s theorem): if f is a real, bounded continuous
function defined on a closed set F ⊂ X, there is a continuous extension of f to
X. Assuming f (F ) ⊂ [1, 2], the case it reduces to, formula
f (x) = d(x, F )
−1 . inf [f (u)d(x, u)] for x ∈ X − F ,
where inf relates to the u ∈ F , provides a solution. Dieudonn´ e, IV, 5. The lemma
correspond to the case F = A ∪ B.
69 i.e. such that every x ∈ X has a compact neighbourhood V . Then any neighbourhood of x contains a compact neighbourhood, for example V ∩ B, where B
is a closed ball centered at x contained in W .