Prove the following.
Theorem. Every topological group is completely regular.
Proof. Let \(V_0\) be a neighborhood of the identity element \(e\), in the topological group \(G\). In general, choose \(V_n\) to be a neighborhood of \(e\) such that \(V_n \cdot V_n \subset V_{n-1}\). Consider the set of all dyadic rationals \(p\), that is, all rational numbers of the form \(k/2^n\), with \(k\) and \(n\) integers. For each dyadic rational \(p\) in \((0, 1]\), define an open set \(U(p)\) inductively as follows: \(U(1) = V_0\) and \(U(\tfrac{1}{2}) = V_1\). Given \(n\), if \(U(k/2^n)\) is defined for \(0 < k/2^n \leq 1\), define
\[ U(1/2^{n+1}) = V_{n+1}, \]
\[ U((2k+1)/2^{n+1}) = V_{n+1} \cdot U(k/2^n) \]
for \(0 < k < 2^n\). For \(p \leq 0\), let \(U(p) = \varnothing\), and for \(p > 1\), let \(U(p) = G\). Show that
\[ V_n \cdot U(k/2^n) \subset U((k+1)/2^n) \]
for all \(k\) and \(n\). Proceed as in the Urysohn lemma.
This exercise is adapted from [M-Z], to which the reader is referred for further results on topological groups.
Solution
1. We must show \(V_n \cdot U(k/2^n) \subset U((k+1)/2^n)\) for all \(k\) and \(n\).
the claim to prove.
2. We prove by induction on \(n\). For \(n = 0\): \(U(0) = \varnothing\) and \(U(1) = V_0\), so \(V_0 \cdot U(0) = \varnothing \subset U(1)\).
base case.
3. Assume the claim holds for \(n\); we prove it for \(n+1\).
induction step.
4. For \(k\) even, \(k = 2j\): \(U(k/2^{n+1}) = U(j/2^n)\), and \(U((k+1)/2^{n+1}) = U((2j+1)/2^{n+1}) = V_{n+1} \cdot U(j/2^n)\).
the definitions.
5. Then \(V_{n+1} \cdot U(k/2^{n+1}) = V_{n+1} \cdot U(j/2^n) = U((k+1)/2^{n+1})\).
4.
6. For \(k\) odd, \(k = 2j+1\): \(U(k/2^{n+1}) = U((2j+1)/2^{n+1}) = V_{n+1} \cdot U(j/2^n)\), and \(U((k+1)/2^{n+1}) = U((2j+2)/2^{n+1}) = U((j+1)/2^n)\).
the definitions.
7. Then \(V_{n+1} \cdot U(k/2^{n+1}) = V_{n+1} \cdot V_{n+1} \cdot U(j/2^n) \subset V_n \cdot U(j/2^n) \subset U((j+1)/2^n) = U((k+1)/2^{n+1})\).
6, using \(V_{n+1} \cdot V_{n+1} \subset V_n\) and the induction hypothesis.
8. Hence \(V_n \cdot U(k/2^n) \subset U((k+1)/2^n)\) for all \(k, n\).
2, 5, 7.
9. Proceeding as in the Urysohn lemma, this defines a continuous function \(f : G \to [0,1]\) with \(f(e) = 0\) and \(f = 1\) outside \(V_0\), separating \(e\) from the closed set \(G \setminus V_0\).
the Urysohn-lemma construction using the nested open sets \(U(p)\).
10. Hence \(G\) is completely regular.
9 (every point can be separated from a closed set by a continuous function, using translation invariance).
11. Q.E.D.
8 and 10.