A a question about the metrization of uniform spaces I have read two theorems about the metrization of uniform spaces from Engelking and Kelley.
Kelley's condition (b) is slightly different from Engelking's corresponding result for Vi's.
I think these two states cannot be obtained from each other. Or, which one is stronger?
Engelking obtained the following 8.1.11 Corollary and 8.1.13 Corollary using 8.1.10. Theorem.
Engelking multiplies the pseudo-metric in 8.1.10 Theorem by 2 while proving 8.1.11 Corollary.
My question is that can we still get the same Corollaries by multiplying the pseudo-metric in Kelley's Theorem by 4?
Kelley says in a different way completely regularity of uniform spaces on page 188 of his book.
I have another question, do the metrization theorems here have any superiority over each other?
In Kelley's book,

In Engelking's book,




 A: The relationship between the two pseudometrics is the following: $\rho(x,y)=2\cdot d(x,y)$ for all $x$ and $y$.
We compare the definitions of the pseudometrics in both books.
Both start with a sequence $\langle U_n:n\in\omega\rangle$ of sets that
contain the diagonal.
The demands on the sequence are: $U_0=X\times X$,
and $U_{n+1}\circ U_{n+1}\circ U_{n+1}\subseteq U_n$ for all $n$.
Engelking specifies that the $U_n$ are members of a uniformity $\mathcal{U}$
and thus, according to his definition, the $U_n$ are, in addition, symmetric.
Kelley imposes no further restrictions.
Kelley defines an auxiliary function $f$ on $X\times X$ by
$$
f(x,y)=\begin{cases} 2^{-n} & (x,y)\in U_{n-1}\setminus U_n\\
                     0     & (x,y)\in \bigcap_n U_n \end{cases}
$$
and defines
$d(x,y)$ to be the infimum of the set of all sums
$$
\sum_{j=0}^n f(x_j,x_{j+1}) 
$$
over all finite sequences $\langle x_i:i\le n+1\rangle$ with $x_0=x$
and $x_{n+1}=y$.
Engelking's definition of $\rho(x,y)$ is a bit different: for every
finite sequence $\langle x_i:i\le n+1\rangle$ of points in $X$ with
$x_0=x$ and $x_{n+1}=y$ he takes all possible sums
$$
\sum_{j=1}^{n+1} 2^{-i_j}
$$
subject to $(x_{j-1},x_j)\in U_{i_j}$.
One can rewrite these sums as
$$
\sum_{j=0}^{n} 2^{-i_j}
$$
subject to $(x_j,x_{j+1})\in U_{i_j}$, this merely involves shifting the sequence
$\langle i_j:1\le j\le n+1\rangle$ down by one.
Now observe that, in general,  $(x,y)\in U_i$ iff $f(x,y)\le 2^{-(i+1)}$
Indeed: if $(x,y)\in U_i$ then $(x,y)\in U_{n-1}\setminus U_n$ for
some $n$ with $n-1\ge i$, that is for some $n>i$.
And that implies that $f(x,y)=2^{-n}\le2^{-(i+1)}$.
Conversely, if $f(x,y)\le2^{-(i+1)}$ then $(x,y)\in U_{n-1}\setminus U_n$
for some $n\ge i+1$, that is, for some $n>i$.
This means that for every sequence $\langle x_i:i\le n+1\rangle$ and every
associated sum $\sum_{j=0}^{n} 2^{-i_j}$ we have
$$
2\cdot \sum_{j=0}^n f(x_j,x_{j+1}) \le 2\cdot\sum_{j=0}^{n} 2^{-(i_j+1)}
                          =\sum_{j=0}^{n} 2^{-i_j}
$$
it also implies that $2\sum_{j=0}^n f(x_j,x_{j+1})$ is one of the associated sums,
namely the one where every $i_j$ is such that
$(x_j,x_{j+1})\in U_{i_j}\setminus U_{i_j+1}$.
This implies that $\rho(x,y)=2\cdot d(x,y)$ for all $x$ and $y$.
