Let \(X\) be a topological space . If there is a connected dense subset \(S\subseteq X\), then \(X\) is also connected.
Proof
Assume for the sake of contradiction that \(U\) and \(V\) disconnect \(X\). Since \(S\) is connected it lies either \(U\) or in \(V\) (see (0x680b53b0)
).
The density of \(S\) implies that \(U=X\), which is a contradiction.