3. SOLUTIONS TO THE PROBLEMS
58
We now prove that the connected component of 0 consists of 0 alone; to prove this, it is enough to show that the intersection of all open-closed sets containing 0 is 0. As p k H is an open subgroup, it is closed as well, and as the intersection of all of them (for k = 1, 2, ...) is already 0 alone, we get the desired result. For compactness we are going to prove the following result: H is compact if and only if the index of pkH in H is finite for every k. Necessity of this condition is easy to establish: the cosets X + pkH(X E H) cover H. If H is compact, then already a finite number of them have to cover H, which is precisely what the condition says. Suppose now that all the subgroups pkH have finite index. We are going to show that if a family of closed subsets of H has the property that any finite number of them have a nonempty intersection, then the whole family has a nonempty intersection. First, we show the following: let UA(A E A) be subsets of the set H such that the intersection of any finite number of them is nonempty.
Suppose that H is the disjoint union of two subsets S and T. Then at least one of the families s n UA(A E A) and T n UA(A E A) inherits the same intersection property. (In particular, either S or T meets all the UA.) Suppose that neither family inherited the property. Then we would have values )\1 i .'2, ... , .\,. and )r+1, )tr+2, , \r+9 such that the intersection of the s n Ua; and also of the T n Ua,,+j would be empty (i = 1, 2, ... , r, j = 1, 2, ... , s). In other words, the sets
tnUa,) n S and (uA+) n T are empty, or equivalently
fl Ua; C T and
(UA+.) C S.
Consequently,
/ c T n s= O,
UA: I (rn+ x=1
a contradiction. We then get the same type of statement for any decomposition of H into a finite number of pairwise disjoint subsets. Let us now consider closed subsets U,\ (A E A) of our topological group H
satisfying the finite intersection condition. Repeatedly applying the above procedure, we see that there exists a sequence 711, 772.... , 77k.... of elements of H for which
,q1+pH
772+p2H2...
1)k+pkH
...
(1)
and such that for any k the intersection of the UA(A E A) with 77k + pkH also satisfies the finite intersection condition; in particular, any UA and r!k +pkH have a common element Xk,A