# Topological Groups and Families of Pseudometrics

The topology on a topological group is generated by a family of pseudometrics. The only proof I know passes through uniform spaces (by which I mean the entourage definition): A topological group has a uniformity and by a theorem of Weil, every uniformity comes from a family of pseudometrics. Is there a direct construction of the pseudometrics that bypasses Weil's theorem?

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Hewitt and Ross, "Abstract Harmonic Analysis, I", Chapter II.8, Theorem 8.2, give a direct construction of a (left-invariant) pseudometric, starting from a sequence $U_{k}$ of symmetric neighborhoods such that $U_{k+1}^2 \subset U_{k}$. In essence, their argument is the same as Weil's, however. –  Theo Buehler Dec 27 '10 at 0:45
Using this you can get the sought family of pseudo-metrics by constructing for each neighborhood $V$ of the identity a decreasing sequence $U_{k}$ of symmetric neighborhoods such that $U_{k+1}^{2} \subset U_{k} \subset V$. –  Theo Buehler Dec 27 '10 at 0:56

Both parts are in some sense classical. If you have Munkres close to hand, then the proof of (1) can be extracted by following the recipe given in exercise 5 on page 237. (Yes, he states it for the $T_0$/Hausdorff case, but the proof goes through without that assumption.)
A space $X$ is completely regular iff its Kolmogorov quotient $X_h$ is completely regular (i.e., Tychonoff -- see here), and a Tychonoff space embeds in a product of copies of the unit interval (Munkres, p. 237, theorem 2.2). In this way, the topology of $X_h$ is generated by a family of metrics, and by pulling back these metrics along the quotient map $X \to X_h$, one gets a generating family of pseudometrics for $X$.