Let X be a set, A a family of real-valued functions on X which contains the constants,
μ
A
{\mu _A}
the weak uniformity generated by A, and
U
(
μ
A
X
)
U({\mu _A}X)
the collection of uniformly continuous functions to the real line R. The problem is how to construct
U
(
μ
A
X
)
U({\mu _A}X)
from A. The main result here is: For A a vector lattice, the collection of suprema of countable, finitely A-equiuniform, order-one subsets of
A
+
{A^ + }
is uniformly dense in
U
(
μ
A
X
)
U({\mu _A}X)
. Two less technical corollaries: If A is a vector lattice (resp., vector space), then the collection of functions which are finitely A-uniform and uniformly locally-A (resp., uniformly locally piecewise-A) is uniformly dense in
U
(
μ
A
X
)
U({\mu _A}X)
. Further, for any A, a finitely A-uniform function is just a composition
F
∘
(
a
1
,
…
,
a
p
)
F \circ ({a_1}, \ldots ,{a_p})
for some
a
1
,
…
,
a
p
∈
A
{a_1}, \ldots ,{a_p} \in A
and F uniformly continuous on the range of
(
a
1
,
…
,
a
p
)
({a_1}, \ldots ,{a_p})
in
R
p
{R^p}
. Thus, such compositions are dense in
U
(
μ
A
X
)
U({\mu _A}X)
. For
B
U
(
μ
A
X
)
BU({\mu _A}X)
, the compositions with
F
∈
B
U
(
R
p
)
F \in BU({R^p})
are dense (B denoting bounded functions). So, in a sense, to know
U
(
μ
A
X
)
U({\mu _A}X)
it suffices to know A and subspaces of the spaces
R
p
{R^p}
, and to know
B
U
(
μ
A
X
)
BU({\mu _A}X)
, A and the spaces
R
p
{R^p}
suffice.