Union of transitively overlapping connected spaces
Exercise 225 (Chapter 32, “Connectedness”):
Prove the following generalization of theorem 40. Let
(
in
) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by
if
have just one equivalence class. Then
is connected.
For clarity, rather than a family I will consider a set
of connected subsets of
. The condition is that the equivalence relation in
generated by the above relation
has only one class.
Let .
Let and
be eventual open subsets of
such that
and
. To show that
is connected, we must show that either
or
is empty.
First, let be an eventual element of
. Since
, the two subsets of
,
and
are disjoint, and their union is
. Furthermore, they are open in
. Since
is connected, this implies that one or the other is empty. If
is empty, then
; if instead
is empty, then
. Hence in all cases an element of
is entirely in
or entirely in
.
Let us consider the relation on
defined by
; in other words,
if and only if both are subsets of
, or both are subsets of
. Clearly, this is an equivalence relation.
Furthermore, if , that is,
, we cannot have
and
, for
and
are disjoint. Hence
and
are either both in
or both in
; that is,
. Hence
is an equivalence relation greater than the equivalence relation generated by
.
If had two or more classes, there would be some element of
in one class and another in another; these two elements would not be equivalent following
. Since
is greater than the equivalence relation generated by
, they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence
too can have only one class, which means that for all
and
in
, either both are in
or both are in
. But this implies that all are in
or all are in
, which in turn implies that their union
is a subset of either
or
, hence that either
or
is empty.