Show that the topologist’s sine curve has dimension 1.
Solution
Let \[ T=\{(x,\sin(1/x)):0<x\le1\}\cup(\{0\}\times[-1,1]) \] be the topologist’s sine curve.
For \(n\ge1\), put \[ A_n=\{(x,\sin(1/x)):1/(n+1)\le x\le1/n\}, \] and let \[ V=\{0\}\times[-1,1]. \] Each \(A_n\) is the continuous image of a compact interval under the embedding \[ x\longmapsto (x,\sin(1/x)), \] so it is an arc, hence homeomorphic to a closed interval and therefore has covering dimension \(1\). The set \(V\) is also an arc, so \(\dim V=1\). Each \(A_n\) and \(V\) is compact, hence closed in the metric space \(T\), and \[ T=V\cup\bigcup_{n\ge1}A_n. \]
By the countable closed-sum theorem for covering dimension, if a metrizable space is a countable union of closed subspaces of covering dimension at most \(m\), then its covering dimension is at most \(m\). Hence \[ \dim T\le1. \]
On the other hand, \(T\) is connected, is \(T_1\) (indeed metric), and contains more than one point. By E-KWBI9, every such space has dimension at least \(1\). Thus \[ \boxed{\dim T=1}. \]