alt="script 풜"/> is equiregulated by hypothesis. Then, for every , there exists a division
fulfilling
for every and
,
.
On the other hand, the sets and
are relatively compact in
for every
, where
. Thus, there is a subsequence of indexes
, with
, for which
and
are also relatively compact sets of
, for every
.
This last statement implies that there exist and
satisfying
Thus, there exists such that
provided . In particular, for
, we have
for .
Take and consider
such that
. Therefore, either
, for some
, in which case, we have
or , for some
, in which case, we have