[2BR]Let \((X,𝜏)\) be a topological space. Given \(A,B⊆ X\), to shorten the formulas we will use the (nonstandard) notation
\(A{\mathbf{i}}B\) to say that \(A,B\) have non-empty intersection,
\(A{\mathbf{d}}B\) to say that they are disjointed, and
\({\mathbf{n}}A\) to say that \(A\) it is not empty.
we recall the definition of connectedness (Chap. 5 Sec. 11 of the notes [ 3 ] or, Chap. 2 in [ 26 ] ).
The space \(X\) is disconnected if it is the disjoint union of two open non-empty sets.
The space \(X\) is connected if it is not disconnected. This may be rewritten in different fashions, as for example
\[ ∀ A,B∈ 𝜏, ( {\mathbf{n}}A ~ ∧~ {\mathbf{n}}B ~ ∧~ X⊆ A∪ B ) ⇒ A{\mathbf{i}}B~ . \]A non-empty subset \(E⊆ X\)is disconnected if it is disconnected with the induced topology; that is, if \(E\) is covered by the union of two open sets, each of which intersects \(E\), but which are disjointed in \(E\); in symbols,
\begin{equation} ∃ A,B∈ 𝜏, E{\mathbf{i}}A ~ ∧~ E{\mathbf{i}}B~ ∧~ E⊆ A∪ B~ ∧~ A∩ B∩ E= ∅~ . \label{eq:E_ sconnesso_ lunga} \end{equation}3Similarly a non-empty set \(E⊆ X\) is connected if it is connected with the induced topology. This may be written as
\begin{equation} ∀ A,B∈ 𝜏, (E{\mathbf{i}}A~ ∧ ~ E{\mathbf{i}}B~ ∧ ~ E⊆ A∪ B )⇒ A∩ B∩ E≠ ∅~ .\label{eq:E_ connesso_ lunga} \end{equation}4or equivalently
\begin{equation} ∀ A,B∈ 𝜏, ( E⊆ A∪ B ∧ A∩ B∩ E= ∅ )⇒ ( E⊆ A \lor E⊆ B )~ .\label{eq:E_ connesso_ lunga_ 2} \end{equation}5