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3.2 Connected Components

3.2.1 Definition of components.

Let \(X\) be a topological space and \(x\in X\). The component of \(x\) in \(X\), denoted \(C\of x\) is the union of all connected subsets of \(X\) that contain \(x\).

Remark.

The components of \(X\) are connected subsets.

3.2.2 Proposition (properties of components).

Let \(X\) be a topological space.

  • 1. The set of components of \(X\) is a partition of \(X\).

  • 2. Each component is closed.

  • 3. Each connected subset of \(X\) is contained in a component of \(X\).

  • Proof. To prove 1. we show that if \(x,y\in X\), then either \(C\of x=C\of y\) or \(C\of x\cap C\of y=\emp \). Let \(x,y\in X\). Suppose that \(C\of x\cap C\of y\neq \emp \) then \(C\of x\cup C\of y\) is connected and contains \(x\) so

    \[ C\of x\cup C\of y\sub C\of x \]

    and consequently

    \[ C\of x\cup C\of y=C\of x. \]

    Similarly,

    \[ C\of x\cup C\of y=C\of y. \]

    Thus \(C\of x=C\of y\) as required.

    Let \(x\in X\). Since \(C\of x\) is connected, it follows that \(\ob {C\of x}\) is connected. Thus \(\ob {C\of x}\sub C\of x\) so \(\ob {C\of x}=C\of x\) and \(C\of x\) is closed.

    If \(A\) is a connected subset of \(X\) and \(A\neq \emp \), then \(A\sub C\of x\), where \(x\in A\).  □

Example.

Let \(\bq \) have the subspace topology. Then no subset of \(\bq \) with at least two points is connected (\(\bq \) contains no nontrivial intervals). Thus singletons are the components of \(\bq \). They are not open in \(\bq \).

Example.

Let \(C\) be the Cantor set. The components of \(C\) are singletons, since \(C\) contains no nontrivial interval.

3.2.3 Totally disconnected space.

A topological space \(X\) is totally disconnected if the components of \(X\) are singletons.

Examples.

The set of rational numbers \(\bq \), the Cantor set \(C\) or the set \(\br \sem \bq \) of irrational numbers are all totally disconnected as topological spaces with the subspace topology inherited from \(\br \).

3.2.4 Quasi-components.

Let \(X\) be a topological space and \(K\sub X\). We say that \(K\) is a quasi-component of \(X\) if:

  • 1. for each separation \(\set {A,B}\) of \(X\), either \(K\sub A\) or \(K\sub B\), and

  • 2. for any \(L\sub X\) with \(K\subnq L\) there is a separation \(\set {A,B}\) of \(X\) with \(L\cap A\neq \emp \) and \(L\cap B\neq \emp \).

Example.

Let

\[ J:=\set {\frac {1}{n}:n\in \bn }\tm \bra {0,1}\cup \set 0\tm \cur {\bra {0,1}\sem \set {\frac {1}{2}}} \]

with the subspace topology inherited from \(\br ^{2}\). Then

\[ J_{1}=\set 0\tm \clo {0,\frac {1}{2}}\qand J_{2}:=\set 0\tm \ocl {\frac {1}{2},1} \]

are components of \(J\), but they are not quasi-components. \(J_{1}\cup J_{2}\) is a quasi-component.

3.2.5 Proposition (properties of quasi-components).

Let \(X\) be a space.

  • 1. Each point belongs to a unique quasi-component of \(X\).

  • 2. The quasi-component containing a point \(x\) is the intersection of all clopen subsets of \(X\) that contain \(x\).

  • 3. Each component of \(X\) is contained in a quasi-component of \(X\).

  • Proof. 1. For \(x\in X\), let \(K\of x\) be the set of all \(y\in X\) such that there are no separation \(\set {A,B}\) of \(X\) with \(x\in A\) and \(y\in B\). Then \(K\of x\) is a quasi-component of \(X\) containing \(x\). Suppose that \(K\) and \(K’\) are quasi-components of \(X\) with \(x\in K\cap K’\). If \(\set {A,B}\) is a separation of \(X\), then either \(x\in A\) or \(x\in B\). If \(x\in A\), then \(K\sub A\) and \(K’\sub A\) so \(K\cup K’\sub A\). If \(x\in B\), then \(K\cup K’\sub B\). Thus \(K\cup K’\) can’t be a proper superset of neither \(K\) nor \(K’\). It follows that

    \[ K=K\cup K’=K’. \]

    2. Let \(K\) be a quasi-component of \(X\) with \(x\in K\). If \(A\sub X\) is clopen with \(x\in A\), then either \(A=X\) or \(\set {A,X\sem A}\) is a separation of \(X\). In either case, \(K\sub A\). Thus if \(\sa \) is the family of all clopen subsets of \(X\) that contain \(x\), then \(K\sub \bigcap \sa \). Let \(\set {A,B}\) be any separation of \(X\). If \(x\in A\), then \(A\in \sa \) so \(\bigcap \sa \sub A\). If \(x\in B\), then \(\bigcap \sa \sub B\). Thus \(K=\bigcap \sa \).

    3. Let \(C\) be a component of \(X\). Let \(x\in C\) and \(K\) be the quasi-component of \(X\) with \(x\in K\). Suppose, for a contradiction, that there is \(y\in C\sem K\). Then there is a separation \(\set {A,B}\) of \(X\) with \(x\in A\) and \(y\in B\). This implies that \(\set {C\cap A,C\cap B}\) is a separation of \(C\), which is a contradiction.  □

Remark.

Any component of \(X\) that is clopen is a quasi-component. In particular, if there are only finitely many components, they are clopen so they are quasi-components.

3.2.6 Exercises.