Let \(X\) be a topological space. A separation of \(X\) is a pair \(\set {A,B}\) of nonempty disjoint open subsets of \(X\) with \(A\cup B=X\).
If \(\set {A,B}\) is a separation of \(X\), then both \(A\) and \(B\) are closed.
A topological space \(X\) is connected if it has no separation, otherwise is it is disconnected.
The Sierpiński space is connected. The infinite space with cofinite topology is connected. Any trivial space is connected. A discrete space with more than one point is disconnected.
A topological space \(X\) is connected if and only if any continuous function from \(X\) to a discrete space is constant.
Proof. Assume that \(X\) is connected and \(Y\) is a discrete space. Suppose, for a contradiction, that \(f:X\to Y\) is continuous and not constant. Then there are \(x,x’\in X\) with \(f\of x\neq f\of {x’}\). Since \(Y\) is discrete, the sets \(U:=\set {f\of x}\) and \(V:=Y\sem \set {f\of x}\) are open. Then \(\set {f^{-1}\bof U,f^{-1}\bof V}\) is a separation of \(X\) and we get a contradiction.
Now assume that any continuous function from \(X\) to a discrete space is constant. Suppose, for a contradiction, that \(\set {A,B}\) is a separation of \(X\). Let \(Y:=\set {0,1}\) have the discrete topology. Define \(f\of x:=0\) for \(x\in A\) and \(f\of x:=1\) for \(x\in B\). Then \(f\) is continuous, but not constant, which is a contradiction. □
A subset \(Y\) of a topological space \(X\) is connected if it is connected as a topological space with the subspace topology.
A subset \(Y\) of a topological space \(X\) is connected if and only if there are no open subsets \(A,B\sub X\) such that
1. \(Y\sub A\cup B\),
2. \(A\cap Y\neq \emp \neq B\cap Y,\) and
3. \(A\cap B\cap Y=\emp \).
A subset \(Y\) of \(\br \) is connected if and only if \(Y\) is an interval.
Proof. Assume that \(Y\) is not an interval. Then there are \(a,b\in Y\) and \(c\in \cur {a,b}\sem Y\). With \(A:=\cur {-\ity ,c}\) and \(B:=\cur {c,\ity }\), the pair \(\set {A\cap Y,B\cap Y}\) is a separation of \(Y\) so \(Y\) is disconnected.
Now assume that \(Y\) is disconnected. Let \(A,B\) be open in \(\br \) and such that \(\set {A\cap Y,B\cap Y}\) is a separation of \(Y\). Let \(a\in A\cap Y\) and \(b\in B\cap Y\). Without loss of generality, we can assume that \(a<b\). Let
\[ c:=\sup \set {x\in A:x<b}. \]
Since \(B\) is open in \(\br \) and since \(b\in B\), it follows that \(c\nin B\). Since \(A\) is open, it follows that \(c\nin A\). Thus \(c\nin Y\) and since \(a,b\in Y\) and \(a<c<b\), it follows that \(Y\) is not an interval. □
Let \(X\) be a topological space and \(A,B\sub X\). We say that \(A\) and \(B\) are separated if \(A\cap \ob B=\emp \) and \(\ob A\cap B=\emp \).
Let \(X\) be a topological space and \(Y\sub X\). Then \(Y\) is connected if and only if \(Y\) is not a union of two nonempty separated subsets of \(X\).
Proof. Assume that \(Y\) is connected. Suppose, for a contradiction, that \(Y=A\cup B\), where \(A,B\) are nonempty and separated subsets of \(X\). Let \(U:=X\sem \ob B\) and \(V:=X\sem \ob A\). Then \(\set {U\cap Y,V\cap Y}\) is a separation of \(Y\), which is a contradiction.
Assume that \(Y\) is disconnected. Let \(U,V\) be open in \(X\) and such that \(\set {A,B}\) is a separation of \(Y\), where \(A:=U\cap Y\) and \(B:=V\cap Y\). Proposition 1.5.7 implies that the closure of \(A\) in \(Y\) is equal to \(\ob A\cap Y\). Since \(A\) is closed in \(Y\), it follows that \(\ob A\cap Y=A\) so \(\ob A\cap B=\emp \). Similarly, \(A\cap \ob B=\emp \) so \(A\) and \(B\) are separated in \(X\). Thus \(Y\) is a union of two nonempty separated subsets \(A\) and \(B\) of \(X\). □
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). If \(X\) is connected, then \(Y\) is connected.
Proof. If \(Y\) is disconnected, then there is a separation \(\set {A,B}\) of \(Y\). If follows that \(\set {f^{-1}\bof A,f^{-1}\bof B}\) is a separation of \(X\). □
Let \(X\) be a connected topological space, \(f:X\to \br \) be continuous and \(a<b<c\) be such that \(a\) and \(c\) are values of \(f\). Then \(b\) is also a value of \(f\).
Proof. Since \(f\bof X\) is a connected subset of \(\br \), it is an interval. □
If \(f:\bra {0,1}\to \bra {0,1}\), then \(f\of t=t\) for some \(t\in \bra {0,1}\).
Proof. Suppose, for a contradiction, that such \(t\) does not exist. Then \(f\of 0>0\) and \(f\of 1<1\). Define \(g:\bra {0,1}\to \br \) be \(g\of x:=f\of x-x\). Then \(g\) is continuous, \(g\of 0>0\) and \(g\of 1<0\) so there is \(t\in \cur {0,1}\) with \(g\of t=0\). Then \(f\of t=t\), which is a contradiction. □
Let \(X\) be a topological space and \(\sa \) be a family of connected subsets of \(X\) such that \(A\cap A’\neq \emp \) for any \(A,A’\in \sa \). Then \(\bigcup \sa \) is a connected subset of \(X\).
Proof. Suppose, for a contradiction, that \(\bigcup \sa \) is disconnected. Let \(\set {U,V}\) be a separation of \(\bigcup \sa \). Then there are \(A,A’\in \sa \) with \(U\cap A\neq \emp \) and \(V\cap A’\neq \emp \). Let \(x\in A\cap A’\). If \(x\in V\), then \(\set {U\cap A,V\cap A}\) is a separation of \(A\), which is a contradiction. Similarly, we get a contradiction when \(x\in U\). □
Let \(X\) be a topological space such that for any \(x,y\in X\) there is a connected \(A\sub X\) with \(x,y\in A\). Then \(X\) is connected.
Proof. Suppose, for a contradiction, that \(X\) is disconnected. Let \(\set {A,B}\) be a separation of \(X\). Let \(a\in A\) and \(b\in B\). If \(C\) is a connected subset of \(X\) with \(a,b\in C\), then \(\set {A\cap C,B\cap C}\) is a separation of \(C\), which is a contradiction. □
Let \(X\) be a topological space, \(C\) be a connected subset of \(X\) and \(\sa \) be a family of connected subsets of \(X\) such that \(C\cap Y\neq \emp \) for any \(Y\in \sa \). Then \(C\cup \bigcup \sa \) is connected.
Proof. Suppose, for a contradiction, that \(C\cup \bigcup \sa \) is disconnected. Let \(\set {A,B}\) be a separation of \(C\cup \bigcup \sa \). Since \(C\) is connected, either \(A\cap C=\emp \) or \(B\cap C=\emp \). Assume \(A\cap C=\emp \). Let \(a\in A\) and \(Y\in \sa \) be such that \(a\in Y\). Since \(C\sub B\) and \(C\cap Y\neq \emp \), it follows that \(B\cap Y\neq \emp \) so \(\set {A\cap Y,B\cap Y}\) is a separation of \(Y\), which is a contradiction. □
Let \(X\) be a topological space and \(A\sub X\) be connected. If \(A\sub B\sub \ob A\), then \(B\) is connected.
Proof. Let \(D\) be a discrete space and \(f:B\to D\) be continuous. Then \(f\res A\) is constant. Let \(d\) be the value of \(f\) on \(A\). Since \(B\) is the closure of \(A\) in \(B\), it follows that
\[ f\bof B\sub \ob {f\bof A}=\ob {\set d}=\set d, \]
so \(f\) is constant. Thus \(B\) is connected. □
Let \(X_{\ga }\) be a topological space for every \(\ga \in A\). Then \(X:=\prod _{\ga \in A}X_{\ga }\) is connected if and only if \(X_{\ga }\) is connected for each \(\ga \in A\).
Proof. If \(X\) is connected, then each \(X_{\ga }\) is connected since \(p_{\ga }:X\to X_{\ga }\) is continuous for each \(\ga \in A\).
Assume that \(X_{\ga }\) is connected for each \(\ga \in A\). Let \(y=\cur {y_{\ga }}_{\ga \in A}\in X\) be fixed and let \(\sa \) be the family of all connected subsets of \(X\) containing \(y\). Then \(\bigcup \sa \) is connected so \(\ob {\bigcup \sa }\) is connected. We will show that \(\ob {\bigcup \sa }=X\). Let \(x=\cur {x_{\ga }}_{\ga \in A}\in X\). To show that \(x\in \ob {\bigcup \sa }\) we will take any basic open nbhd \(B\) of \(x\) and prove that there is \(Y\in \sa \) such that \(B\cap Y\neq \emp \), that is, that there is a connected subset \(Y\) of \(X\) with \(y\in Y\) and \(Y\cap B\neq \emp \).
Let \(A’\sub A\) be finite and
\[ B:=\prod _{\ga \in A}B_{\ga }, \]
where \(B_{\ga }\) is an open nbhd of \(x_{\ga }\) for \(\ga \in A’\) and \(B_{\ga }=X_{\ga }\) for \(\ga \in A\sem A’\). Assume that
\[ A’:=\set {\ga _{1},\ga _{2},\ds ,\ga _{n}}. \]
Let
\[ Y_{1}:=\set {\cur {z_{\ga }}_{\ga \in A}:z_{\ga }=y_{\ga }\text { for every }\ga \in A\sem \set {\ga _{1}}}. \]
Then \(Y_{1}\) is homeomorphic to \(X_{\ga _{1}}\) and \(y\in Y_{1}\). Let
\[ Y_{2}:=\set {\cur {z_{\ga }}_{\ga \in A}:z_{\ga _{1}}=x_{\ga _{1}},\,z_{\ga }=y_{\ga }\text { for every }\ga \in A\sem \set {\ga _{1},\ga _{2}}}. \]
Then \(Y_{2}\) is homeomorphic to \(X_{\ga _{2}}\) and \(Y_{1}\cap Y_{2}\neq \emp \). Thus \(Y_{1}\cup Y_{2}\) is connected. By induction, for \(k\in \set {2,3,\ds ,n}\) let
\( \seteqsection {3} \)\begin{align*} Y_{k} & :=\left \{ \cur {z_{\ga }}_{\ga \in A}:z_{\ga _{i}}=x_{\ga _{i}}\text { for }i=1,\ds ,k-1\right .\\ & \qquad \qquad \qquad \left .\text { and }z_{\ga }=y_{\ga }\text { for }\ga \in A\sem \set {\ga _{1},\ds ,\ga _{k}}\right \} . \end{align*} Then \(Y_{k}\) is homeomorphic to \(X_{\ga _{k}}\) and \(Y_{k-1}\cap Y_{k}\neq \emp \). Thus
\[ \cur {Y_{1}\cup \ds \cup Y_{k-1}}\cap Y_{k}\neq \emp \]
so \(Y_{1}\cup \ds \cup Y_{k}\) is connected. In particular,
\[ Y:=Y_{1}\cup Y_{2}\cup \ds \cup Y_{n} \]
is connected and \(y\in Y\) since \(y\in Y_{1}\). Moreover, \(Y\cap B\neq \emp \) since \(x\in Y_{n}\). □