<P> This means that, if the union X (\ displaystyle X) is disconnected, then the collection (X i) (\ displaystyle \ (X_ (i) \)) can be partitioned to two sub-collections, such that the unions of the sub-collections are disjoint and open in X (\ displaystyle X) (see picture). This implies that in several cases, a union of connected sets is necessarily connected . In particular: </P> <Ol> <Li> If the common intersection of all sets is not empty (∩ X i ≠ ∅ (\ displaystyle \ cap X_ (i) \ neq \ emptyset)), then obviously they cannot be partitioned to collections with disjoint unions . Hence the union of connected sets with non-empty intersection is connected . </Li> <Li> If the intersection of each pair of sets is not empty (∀ i, j: X i ∩ X j ≠ ∅ (\ displaystyle \ forall i, j: X_ (i) \ cap X_ (j) \ neq \ emptyset)) then again they cannot be partitioned to collections with disjoint unions, so their union must be connected . </Li> <Li> If the sets can be ordered as a "linked chain", i.e. indexed by integer indices and ∀ i: X i ∩ X i + 1 ≠ ∅ (\ displaystyle \ forall i: X_ (i) \ cap X_ (i + 1) \ neq \ emptyset), then again their union must be connected . </Li> <Li> If the sets are pairwise - disjoint and the quotient space X / (X i) (\ displaystyle X / \ (X_ (i) \)) is connected, then X (\ displaystyle X) must be connected . Otherwise, if U ∪ V (\ displaystyle U \ cup V) is a separation of X (\ displaystyle X) then q (U) ∪ q (V) (\ displaystyle q (U) \ cup q (V)) is a separation of the quotient space (since q (U), q (V) (\ displaystyle q (U), q (V)) are disjoint and open in the quotient space). </Li> </Ol> <Li> If the common intersection of all sets is not empty (∩ X i ≠ ∅ (\ displaystyle \ cap X_ (i) \ neq \ emptyset)), then obviously they cannot be partitioned to collections with disjoint unions . Hence the union of connected sets with non-empty intersection is connected . </Li> <Li> If the intersection of each pair of sets is not empty (∀ i, j: X i ∩ X j ≠ ∅ (\ displaystyle \ forall i, j: X_ (i) \ cap X_ (j) \ neq \ emptyset)) then again they cannot be partitioned to collections with disjoint unions, so their union must be connected . </Li>

Show that any set obtained by removing a single point from r2 is still connected