Let \(X\) be a topological space. Let \(A\subset X\) be connected. Prove that the closure \(\overline A\) of \(A\) is connected.
Solution
1. Regard \(A\) as a subspace of \(\overline A\).It is connected with this subspace topology.
The topology that \(A\) inherits from \(\overline A\) is the same as the topology that it inherits directly from \(X\): if \(U\) is open in \(X\), then \[ A\cap(U\cap\overline A)=A\cap U, \] because \(A\subseteq\overline A\). Thus the given connectedness of \(A\) is unchanged when \(A\) is viewed as a subspace of \(\overline A\).
2. Suppose, for contradiction, that \(\overline A\) is disconnected.Then there are nonempty disjoint subsets \(U,V\subseteq\overline A\) that are open in \(\overline A\) and satisfy\[ \overline A=U\cup V. \]
This is exactly the definition of a separation of the subspace \(\overline A\).
3. The connected set \(A\) is contained entirely in one of \(U\) or \(V\).
1. Hence one is empty.After interchanging \(U\) and \(V\) if necessary, assume\[ A\subseteq U. \]
4. The set \(U\) is closed in \(\overline A\).
Since \[ \overline A\setminus U=V \] and \(V\) is open in \(\overline A\), the set \(U\) is closed in \(\overline A\).
5. The closure of \(A\) taken inside the subspace \(\overline A\) is all of \(\overline A\).
The closure of a subset \(B\subseteq Z\subseteq X\) in the subspace \(Z\) is \[ \operatorname{Cl}_Z(B)=Z\cap\operatorname{Cl}_X(B). \] Taking \(B=A\) and \(Z=\overline A\) gives \[ \operatorname{Cl}_{\overline A}(A) =\overline A\cap\operatorname{Cl}_X(A) =\overline A\cap\overline A =\overline A. \]
6. The assumed separation is impossible.
By 3, \(A\subseteq U\). By 4, \(U\) is closed in \(\overline A\). Therefore \(U\) contains the closure of \(A\) in \(\overline A\). By 5, that closure is all of \(\overline A\), so \[ \overline A\subseteq U. \] Hence \(U=\overline A\), forcing \(V=\varnothing\), contrary to 2.
7. Therefore \(\overline A\) is connected.
6. Thus no separation exists, so \(\overline A\) is connected.