Connectedness

• $X$,
  • topological sp.

Definiton disjoint set

• $A$
  • set
• $B_{1} \subseteq A$,
• $B_{2} \subseteq A$,

$B_{1}, B_{2} $ is said to be disjoint set of $A$ if

\[B_{1} \cap B_{2} = \emptyset, \ B_{1} \cup B_{2} = A.\]

We write $B_{1} \sqcup B_{2} = A$.

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Definition disconnected

$X$ is said to be disconnected if

\[\exists O_{1}, O_{2} \in \mathcal{O} \text{ s.t. } O_{1} \sqcap O_{2} = X, .\]

$X$ is said to be connected if

\[\forall O_{1}, O_{2} \in \mathcal{O}, \ O_{1} \cap O_{2} = \emptyset \Rightarrow O_{1} \cup O_{2} \neq X .\]

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Definition separated

• $A \subseteq X$,
• $B \subseteq X$,

$X$ is said to be disconnected if

\[\left( B \cap \mathrm{cl}_{X}A \right) \cup \left( A \cap \mathrm{cl}_{X}B \right) = \emptyset .\]

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Theorem 1

• $X$,
  • toplogical sp.
• (1) $\emptyset$ and $X$ are the only closed and open sets in $X$,
• (3)

\[\forall O_{1}, O_{2} \in \mathcal{O}_{X}, \ O_{1} \neq \emptyset, \ O_{2} \neq \emptyset, \Rightarrow O_{1} \sqcup O_{2} \neq X\]

• (4)

\[\forall O_{1}, O_{2} \in \mathcal{O}_{X}, \ O_{1} \neq \emptyset, \ O_{2} \neq \emptyset, \ O_{1}^{c} \sqcup O_{2}^{c} \neq X\]

(5) $X$ is not the union of two nonempty separated sets

proof

(1) $\Rightarrow$ (3)

Suppose that (3) is false. There are \(O_{1}, O_{2} \in \mathcal{O}_{X}\) such that \(O_{1} \neq \emptyset$, $O_{2} \neq \emptyset\), \(O_{1} \sqcup O_{2} = X\). However, since, \(X \setminus O_{1} = O_{2} = O_{1}^{c}\), $O_{2}$ is closed. Thus, $O_{2}$ is closed and open, but this contradicts to (1).

(3) $\Leftrightarrow$ (4)

Suppose that (4) is false. There are \(O_{1}, O_{2} \in \mathcal{O}_{X}\) such that \(O_{1}^{c} \neq \emptyset\), \(O_{2}^{c} \neq \emptyset\), \(O_{1}^{c} \sqcup O_{2}^{c} = X\).

\[O_{1}^{c} = X \setminus O_{2}^{c}, \ O_{2}^{c} = X \setminus O_{1}^{c} .\]

Hence \(O_{1}^{c}\) and \(O_{2}^{c}\) are also open. Since \(O_{1}^{c} \sqcup O_{2}^{c} = X\), this contradicts to (3).

The inverse is obvious from this discussion.

(4) $\Rightarrow$ (5)

Supposet that (5) is false. There are separated sets $A, B$ such that $A \neq \emptyset$, $B \neq \emptyset$ $A \cup B = X$. Since \(\mathrm{cl}_{X}B \cap A = \emptyset\) and \(\mathrm{cl}_{X}B \subseteq X\), \(\mathrm{cl}_{X}B \subseteq B\). Hence $B$ is closed. Similarly, $A$ is closed. Since $A$, $B$ is closed, this contradicts to (4).

(5) $\Rightarrow$ (1)

Suppose that (1) is false and $A$ is proper subset of $X$ and closed and open. Since $B := X \setminus A$ is nonempty, closed and open, $\mathrm{cl}A = A$ and $\mathrm{cl}B = B$. It follows that $(A \cap \mathrm{cl}B) \cup (\mathrm{cl}A \cap B) = \emptyset$. Thus, $A$ and $B$ are separated sets.

$\Box$

Definition

• $X$,
• $I := [0, 1]$,
• $p, q \in X$,
• $f:I \rightarrow X$,

$f$ is said to be a path in $X$ from p to q if

• $f(0) = p$,
• $f(1) = q$,
• $f$ is continuous.

$X$ is said to be path-connected if for every point $p, q \in X$ there exists a path in $X$ from $p$ to $q$.

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Definition path-connected

• $X$,
  • top sp

$X$ is said to be path-connected if for any points $a, b \in X$, there is a continuous map $\gamma: [0, 1] \rightarrow X$ such that

\[\gamma(0) = a, \ \gamma(1) = b.\]

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Reference

• Chapter 5 Connectedness