Separable locally compact metric spaces
Lemma. Assume (X, d) is a separable locally compact metric space. Then there exists an
equivalent metric d0 on X (meaning that the identity map (X, d) → (X, d0 ) is a homeomorphism) such that every bounded subset of (X, d0 ) is relatively compact. Furthermore, for any
such d0 the metric space (X, d0 ) is complete.
Proof. To define d0 we will construct a proper function f : X → [0, +∞), that is, a continuous
function such that f −1 ([0, n]) is compact for every n ∈ N. For this, chose a sequence {Un }∞
n=1
of open subsets of X such that U = ∪n:Un ⊂U Un for any open set U ⊂ X. Omitting all Un
that are not relatively compact we may assume that Un is relatively compact for every n.
Then by induction construct compact subsets Kn ⊂ X as follows. Put K1 = U1 . If Kn−1 is
mn
n
constructed, choose mn ≥ n such that Kn−1 ⊂ ∪m
k=1 Uk . Then put Kn = ∪k=1 Uk .
We thus get an increasing sequence of compact sets such that ∪∞
n=1 Kn = X and for every
n the set Kn is contained in the interior of Kn+1 . Now for every n ≥ 1 choose a continuous
function fn such that 0 ≤ fn ≤ 1, fn (x) = 0 for x ∈ Kn−1 , fn (x) = 1 for x outside the interior
of Kn . Finally, put
∞
X
fn (x).
f (x) =
n=1
Then, if f (x) ≤ n, the point x must be contained in the interior of Kn+1 (otherwise it would
be outside the interior of Km for all m ≤ n + 1 and therefore f (x) ≥ n + 1), hence f −1 ([0, n])
is a closed subset of Kn+1 , so it is compact.
Given a proper function f : X → [0, +∞) we can define d0 by
d0 (x, y) = d(x, y) + |f (x) − f (y)|.
Since d ≤ d0 , the identity map (X, d0 ) → (X, d) is continuos. Since f is continuous (with
respect to the metric d), the identity map (X, d) → (X, d0 ) is continuos as well. For x ∈ X
and n ≥ 1 the ball with center x of radius n with respect to the metric d0 is contained in
f −1 ([f (x) − n, f (x) + n]), hence it is relatively compact. Thus d0 has the required properties.
Finally, assume d0 is any metric equivalent to d such that every bounded subset of (X, d0 )
is relatively compact. Assume {xn }n is a Cauchy sequence in (X, d0 ). Then the closed set
K = {xn }n is bounded in (X, d0 ), hence it is compact. It follows that {xn }n converges to a
point in K. Therefore (X, d0 ) is complete.
A useful way of thinking of separable locally compact metric spaces is as follows. Assume X
is such a space and assume that X is not compact. Define a new space X̄, called the one-point
compactification of X, by formally adding a point ∞ to X. We want to define a metric d¯ on
X̄ such that the inclusion map X → X̄ = X ∪ {∞} is continuous and every sequence in X
that eventually leaves every compact subset X converges to ∞ ∈ X̄. It is not very difficult
¯ is compact, the metric d¯ is unique up to
to show that if such a metric d¯ exists, then (X̄, d)
equivalence and its restriction to X is equivalent to the metric we started with. In order to
prove existence, choose a dense sequence {fn }n in the unit ball of Cc (X) with respect to the
supremum norm (the separability of Cc (X) follows from separability of C(Kn ), where Kn are
as in the proof above). We can think of fn as functions on X̄ by letting fn (∞) = 0. Then,
for x, y ∈ X̄, put
∞
X
¯ y) =
d(x,
2−n |fn (x) − fn (y)|.
n=1
1
2
It is easy to see that d¯ has the required properties. Thus, up to replacing a metric by an
equivalent one, a separable locally compact metric space is just a compact metric space with
one point removed.
Note that by compactness of X̄ the complement of any neighborhood of ∞ is a relatively
¯ ∞)−1 is a proper function on X, something we
compact subset of X. Therefore x 7→ d(x,
struggled to construct in the proof above.