Editing Connected space (section)

== Set operations ==
[[File:Union et intersection d'ensembles.svg|thumb|Examples of unions and intersections of connected sets]]The [[intersection]] of connected sets is not necessarily connected.

The [[union (set theory)|union]] of connected sets is not necessarily connected, as can be seen by considering <math>X=(0,1) \cup (1,2)</math>.

Each ellipse is a connected set, but the union is not connected, since it can be partitioned into two disjoint open sets <math>U</math> and <math>V</math>.

This means that, if the union <math>X</math> is disconnected, then the collection <math>\{X_i\}</math> can be partitioned into two sub-collections, such that the unions of the sub-collections are disjoint and open in <math>X</math> (see picture). This implies that in several cases, a union of connected sets {{em|is}} necessarily connected. In particular:

# If the common intersection of all sets is not empty (<math display="inline"> \bigcap X_i \neq \emptyset</math>), then obviously they cannot be partitioned to collections with [[disjoint union]]s. Hence the union of connected sets with non-empty intersection is connected.
# If the intersection of each pair of sets is not empty (<math>\forall i,j: X_i \cap X_j \neq \emptyset</math>) then again they cannot be partitioned to collections with disjoint unions, so their union must be connected.
# If the sets can be ordered as a "linked chain", i.e. indexed by integer indices and <math>\forall i: X_i \cap X_{i+1} \neq \emptyset</math>, then again their union must be connected.
# If the sets are pairwise-disjoint and the [[Quotient space (topology)|quotient space]] <math>X / \{X_i\}</math> is connected, then {{mvar|X}} must be connected. Otherwise, if <math>U \cup V</math> is a separation of {{mvar|X}} then <math>q(U) \cup q(V)</math> is a separation of the quotient space (since <math>q(U), q(V)</math> are disjoint and open in the quotient space).<ref>{{cite web |last1=Eck |first1=David J. |title=Connected Sets |url=https://math.hws.edu/eck/math331/f19/4-connected.pdf |website=Department of Mathematics and Computer Science |publisher=Hobart and William Smith Colleges |access-date=17 March 2025 |archive-url=https://web.archive.org/web/20240901004123/https://math.hws.edu/eck/math331/f19/4-connected.pdf |archive-date=1 September 2024 |url-status=live}}</ref>

The [[set difference]] of connected sets is not necessarily connected. However, if <math>X \supseteq Y</math> and their difference <math>X \setminus Y</math> is disconnected (and thus can be written as a union of two open sets <math>X_1</math> and <math>X_2</math>), then the union of <math>Y</math> with each such component is connected (i.e. <math>Y \cup X_{i}</math> is connected for all <math>i</math>).

{{math proof|title=Proof<ref>{{cite web |author=Marek |title=How to prove this result about connectedness? |date=February 13, 2013 |work=[[Stack Exchange]] |url=https://math.stackexchange.com/q/302094 }}</ref>{{Better source needed|date=December 2024}}|proof=
By contradiction, suppose <math>Y \cup X_{1}</math> is not connected. So it can be written as the union of two disjoint open sets, e.g. <math>Y \cup X_{1}=Z_{1} \cup Z_{2}</math>. Because <math>Y</math> is connected, it must be entirely contained in one of these components, say <math>Z_1</math>, and thus <math>Z_2</math> is contained in <math>X_1</math>. Now we know that:
<math display="block">X=\left(Y \cup X_{1}\right) \cup X_{2}=\left(Z_{1} \cup Z_{2}\right) \cup X_{2}=\left(Z_{1} \cup X_{2}\right) \cup\left(Z_{2} \cap X_{1}\right)</math>
The two sets in the last union are disjoint and open in <math>X</math>, so there is a separation of <math>X</math>, contradicting the fact that <math>X</math> is connected.
}}

[[File:Connectedness-of-set-difference.png|thumb|Two connected sets whose difference is not connected]]