8.5 Connection[2BG]

Definition 253

[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 [ 2 ] or, Chap. 2 in [ 22 ] ).

  • 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}
    254

  • Similarly 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}
    255

    or 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}
    256

Remark 257

[2BS]It is customary to assume that the empty set is connected; this case, however, is of little interest, generally we will exclude it in the following exercises.

There are many equivalent ways of expressing the above definitions; we leave them as (simple) exercises. This Lemma may also be useful.

Lemma 258

[2FY]If \(Y⊆ X\) is connected and \(Y⊆ E ⊆ \overline Y\), then \(E\) is connected.

For the proof, See Teorema 5.11.6 in [ 2 ] , or Theorem 20 in Cap. 1 in [ 14 ] .

E258

[2BT]Show that the assertions ??,?? in 253 are equivalent. Hidden solution: [UNACCESSIBLE UUID ’2BV’]

E258

[0JF]The space \(X\) is disconnected if and only if it is the disjoint union of two non-empty closed sets.

E258

[0JG]A non-empty subset \(E⊆ X\) is disconnected if \(E\) is covered by the union of two closed sets, each of which intersects \(E\), but which are disjoint inside \(E\).

E258

[0JH]Prerequisites:253.\(X\) is disconnected if and only if there exist non-empty sets \(A,B⊆ X\) whose union covers \(X\), but such that \(\overline B {\mathbf{d}}A\) and \( B {\mathbf{d}}\overline A\).

Hidden solution: [UNACCESSIBLE UUID ’0JJ’]

E258

[0JK] Difficulty:*.Suppose \(E⊆ X\) is disconnected, can we assume that

\begin{equation} ∃ A,B∈ 𝜏, E{\mathbf{i}}A ~ ∧~ E{\mathbf{i}}B~ ∧~ E⊆ A∪ B~ ∧~ A{\mathbf{d}}B~ .\label{eq:E_ sconnesso_ disgiunti} \end{equation}
259

that is, that there exist two disjoint open sets, each of which intersects \(E\), and that \(E\) is covered by their union?

Hidden solution: [UNACCESSIBLE UUID ’0JM’][UNACCESSIBLE UUID ’0JP’] See also 4.

E258

[2DK]Let \((X,𝜏_ X)\) be a topological space, \(Y⊆ X\) the topological space with the induced topology

\[ 𝜏_ Y = \{ A∩ Y: A∈ 𝜏_ X\} ~ . \]

Fix \(E⊆ Y\), consider these statements.

(cX)

\(E\) is a connected set in the topological space \((X,𝜏_ X)\);

(cY)

\(E\) is a connected set in the topological space \((Y,𝜏_ Y)\).

Are the two statements equivalent?

Hidden solution: [UNACCESSIBLE UUID ’113’]

E258

[2BW]Note:Proposition 5.11.2 notes [ 2 ] .

A set \(E\subseteq X\) is disconnected if and only there exists a continuous function \(f:E\to {\mathbb {R}}\) that assumes exactly two values, for example \(f(E)=\{ 0,1\} \).

E258

[0JQ] Note:Theorem 5.11.7 notes [ 2 ] .

Let \(I\) be a family of indices. Show that if \(E_ i\) is a family of connected subsets of \(X\) such that

\[ ∀ {i,j∈ I}~ ,~ E_ i∩ E_ j≠∅~ , \]

then \(E=⋃_{i\in I} E_ i\) is connected.

Hidden solution: [UNACCESSIBLE UUID ’0JR’] [UNACCESSIBLE UUID ’0JS’]

Definition 260

[0JT] Given \(x∈ X\), we will say that the connected component of \(X\) containing \(x\) is the union of all the connected sets that contain \(x\) (note that the singleton \(\{ x\} \) is connected). The previous exercise 8 shows that the connected component is connected.

E260

[0JV] Note:Section 5.11.2 in the text [ 2 ] .Show that two connected components are either disjoint or coincide. So the space \(X\) is partitioned into connected components.

Hidden solution: [UNACCESSIBLE UUID ’0JW’]

[UNACCESSIBLE UUID ’0JX’]

[0JY] Let \(C⊆ X\) be a closed set; let \(K\) be a connected component of \(C\): show that \(K\) is closed. Hidden solution: [UNACCESSIBLE UUID ’0JZ’] [2FZ]Find an example of a space \((X,\tau )\) where there is a connected component that is not open. Hidden solution: [UNACCESSIBLE UUID ’2G0’] [0K0] Let now \((X,d)\) be a metric space where open balls \(B(x,r)\) are also closed. Show that the connected components of \(X\) are all and only singletons \(\{ x\} \).

(A space where connected sets are always singletons, is called totally disconnected).

Hidden solution: [UNACCESSIBLE UUID ’0K1’]

[UNACCESSIBLE UUID ’0K3’] See also the exercises in Sec. 10.5.