5.2.1. Connectedness of unions of sets
Lemma 3 If X ∪ Y = A ∪ B and X, Y 6= ∅ and X ∩ Y = ∅ then either
{X, Y } = {A, B} or A intersects both X and Y or B intersects both X and Y
(for every sets A, B, X, Y ).
Proof Let {X, Y } 6= {A, B}. Suppos e that “A inters ects both X and Y ” do es
not hold (for example suppose that A ∩ X = 0) and prove “B intersects both
X and Y ”.
We have X ⊆ B and thus B ∩ X 6= 0. If also B ∩ Y = 0 then B ⊆ X. So
X = B and thus either Y = A what contradicts to our supposition or A ⊃ Y in
which case A intersects both X and Y .
Theorem 7 If sets A, B ∈ PU are connected regarding an exten dable connec-
tor space (U; r) and A r B then A ∪ B is also connected regarding (U; r).
Proof We need to prove that
∀X, Y ∈ P(A ∪ B) \ {∅} : (X ∪ Y = A ∪ B ∧ X ∩ Y = ∅ ⇒ X r Y ).
Let X, Y ∈ P(A ∪ B) \ {∅} and X ∪ Y = A ∪ B ∧ X ∩ Y = ∅. Then by the
lemma either {X, Y } = {A, B} and thus X r Y ⇔ A r B so having X r Y , or A
inters ects both X and Y or B intersects b oth X and Y . Consider for example
then case X ∩ A 6= ∅ and Y ∩ A 6= ∅.
In this case we have (X ∩ A) ∪ (Y ∩ A) = (X ∪ Y ) ∩ A = (A ∪ B) ∩ A = A
and (X ∩ A) ∩ (Y ∩ A) ⊆ X ∩ Y = ∅. Thus X ∩ A r Y ∩ A and consequently
X r Y (taken in account extendability).
Corollary 3 If sets A, B ∈ PU are connected regarding an extendable connec-
tor space (U; r) and A ∩ B 6= ∅ then A ∪ B is also connected regarding (U ; r ).
Proof Replace r with its normalization N(r). This preserves the same con-
nectedness. A ∩ B 6= ∅ ⇒ A N (r) B. Thus we can apply the theorem.
There holds also inﬁnite version of the previous corollary:
Theorem 8 If S ∈ PP U is a collection of connected (regarding an extendable
connector space (U; r)) sets and
T
S 6= ∅ then
S
S is connected (regarding this
connector space).
Proof Let {X, Y } is a partition of
S
S. Then exist a point p ∈
T
S such
that p ∈ X or p ∈ Y . Without lost of generality we may assume p ∈ X. Since
Y 6= ∅, we have q ∈ Y for some q ∈
S
S that is q ∈ A for some A ∈ S. So
A ∩ X, A ∩ Y 6= ∅ and thus {A ∩ X, A ∩ Y } is a partition of A. Since A is
connected, we have A ∩ X r A ∩ Y and thus (taken in account extendability)
X r Y . So
S
S is connected.
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