Show that any discrete space has dimension 0.
Solution
Let \(\mathcal U\) be an open cover of a discrete space \(X\). The family of singletons \[ \mathcal V=\{\{x\}:x\in X\} \] is an open cover refining \(\mathcal U\): for each \(x\), choose \(U_x\in\mathcal U\) containing \(x\), and then \(\{x\}\subset U_x\). No point belongs to more than one member of \(\mathcal V\), so \(\mathcal V\) has order \(1\). Thus \(\dim X\le0\). For a nonempty space the dimension cannot be less than \(0\), hence \(\dim X=0\). (The empty space has the conventional dimension \(-1\).)