Show that the ordered square is locally connected but not locally path connected. What are the path components of this space?
Solution
Write \(I_o^2\) for \([0,1]\times[0,1]\) in the dictionary order topology and \((x,y)\) for \(x\times y\). The ordered set \(I_o^2\) is a linear continuum: its order is dense and it has the least upper bound property.
1. \(I_o^2\) is locally connected.
Every point has a neighborhood basis of open intervals, together with the intervals \([(0,0),q)\) and \((q,(1,1)]\) at the endpoints. These are convex subsets of the linear continuum \(I_o^2\), hence connected.
2. Each vertical fiber \(\{x\}\times[0,1]\) is path connected.
The fiber is the closed interval \([(x,0),(x,1)]\), whose subspace topology is its order topology, and \(t\mapsto(x,t)\) is an order isomorphism from \([0,1]\) onto it, hence a homeomorphism.
3. A path in \(I_o^2\) lies in a single vertical fiber.
Let \(\gamma\colon[0,1]\to I_o^2\) be a path from \((x,y)\) to \((x',y')\) with \(x<x'\). Its image is connected, so it contains the interval between its endpoints, and in particular the point \((t,\frac12)\) for every \(t\in(x,x')\). The sets \(\{t\}\times(0,1)\) for \(x<t<x'\) are uncountably many pairwise disjoint open sets in \(I_o^2\), and their preimages under \(\gamma\) are pairwise disjoint nonempty open subsets of \([0,1]\). Each contains a rational number, which is impossible for uncountably many disjoint sets.
4. The path components of \(I_o^2\) are the vertical fibers \(\boxed{\{x\}\times[0,1]}\), \(0\le x\le1\).
Steps 2 and 3.
5. \(I_o^2\) is not locally path connected.
Let \(0<x\le1\). Every neighborhood of \((x,0)\) contains an interval \(((a,b),(x,0)]\) with \(a<x\), hence points \((t,\frac12)\) with \(a<t<x\). By step 3 such a point is not joined to \((x,0)\) by a path, so no neighborhood of \((x,0)\) is path connected.
6. Q.E.D.
Steps 1, 5, and 4.