 In ordinary analysis the continuum \(\bR\) is connected in the sense that it cannot be split into two non empty subsets neither of which contains a limit point of the other. In smooth infinitesimal analysis it has the vastly stronger property of indecomposability: it cannot be split in any way whatsoever into two disjoint nonempty subsets. For suppose \(\bR = U \cup V\) with \(U \cap V = \varnothing\). Define \(f: \bR \rightarrow \{0, 1\}\) by \(f(x) = 1\) if \(x \in U, f(x) = 0\) if \(x \in V\). We claim that \(f\) is constant. For we have