Prove that \(\mathbb R\) with the usual topology is connected.
Solution
1. Suppose, for contradiction, that\[ \mathbb R=U\cup V \]is a separation, with \(U\) and \(V\) disjoint nonempty open subsets of \(\mathbb R\).Choose\[ u\in U, \qquad v\in V. \]After interchanging \(u\) and \(v\) and simultaneously interchanging \(U\) and \(V\) if necessary, assume\[ u<v. \]
If \(u>v\), rename the point in \(V\) as the left endpoint and the point in \(U\) as the right endpoint, and interchange the names of the two sets. The hypotheses of a separation are symmetric in \(U\) and \(V\).
2. The set\[ S=U\cap[u,v] \]is nonempty and bounded above, so it has a supremum\[ c=\sup S \]with\[ u\le c\le v. \]
Since \(u\in U\cap[u,v]\), the set \(S\) is nonempty. Every element of \(S\) lies in \([u,v]\), so \(v\) is an upper bound. The least-upper-bound property of \(\mathbb R\) therefore gives \(c=\sup S\). Since \(u\in S\), we have \(u\le c\), and since \(v\) is an upper bound, \(c\le v\).
3. The point \(c\) cannot lie in \(U\).
Suppose \(c\in U\). Because \(v\in V\) and \(U\cap V=\varnothing\), we have \(c\ne v\), so \[ c<v. \] Since \(U\) is open, choose \(\varepsilon>0\) such that \[ (c-\varepsilon,c+\varepsilon)\subseteq U. \] Choose \[ t\in\left(c,\min\{c+\varepsilon,v\}\right). \] Then \[ t\in U\cap[u,v]=S \] and \(t>c\), contradicting that \(c\) is an upper bound of \(S\). Hence \(c\notin U\).
4. The point \(c\) cannot lie in \(V\).
Suppose \(c\in V\). Because \(u\in U\) and \(U\cap V=\varnothing\), we have \(c\ne u\), so \[ u<c. \] Since \(V\) is open, choose \(\varepsilon>0\) such that \[ (c-\varepsilon,c+\varepsilon)\subseteq V. \] Set \[ \delta=\min\left\{\frac\varepsilon2,\frac{c-u}{2}\right\}>0. \] By the defining property of the supremum, \(c-\delta\) is not an upper bound for \(S\). Hence there exists \(s\in S\) with \[ c-\delta<s\le c. \] By 3, \(c\notin U\). Since \(s\in S\subseteq U\), we have \(s\ne c\), and therefore \(s<c\). Also \[ s>c-\delta\ge c-\frac\varepsilon2>c-\varepsilon, \] so \(s\in(c-\varepsilon,c+\varepsilon)\subseteq V\). Thus \(s\in U\cap V\), contradicting disjointness. Therefore \(c\notin V\).
5. Therefore \(\mathbb R\) is connected.
By <1>3–<1>4, \[ c\notin U\cup V, \] contradicting the assumed equality \[ \mathbb R=U\cup V. \] Thus no separation of \(\mathbb R\) exists, so \(\mathbb R\) is connected.