A space \((X,\mathcal T)\) is called locally path-connected if for every \(p\in X\), every open neighborhood of \(p\) contains a path-connected open neighborhood of \(p\). Show that the product of two locally path-connected spaces is locally path-connected.
Solution
Let \(X\) and \(Y\) be locally path-connected spaces.
1. Let \((x_0,y_0)\in X\times Y\) and let \(W\) be an open neighborhood of \((x_0,y_0)\).There are open sets \(U\subseteq X\) and \(V\subseteq Y\) such that\[ (x_0,y_0)\in U\times V\subseteq W. \]
The products of open subsets of \(X\) and \(Y\) form a basis for the product topology. Since \(W\) is open and contains \((x_0,y_0)\), some basic product neighborhood \(U\times V\) of that point is contained in \(W\).
2. There are path-connected open sets \(U_0\subseteq X\) and \(V_0\subseteq Y\) satisfying\[ x_0\in U_0\subseteq U, \qquad y_0\in V_0\subseteq V. \]
Apply local path-connectedness of \(X\) to the open neighborhood \(U\) of \(x_0\), and local path-connectedness of \(Y\) to the open neighborhood \(V\) of \(y_0\).
3. The product \(U_0\times V_0\) is path-connected.
Take arbitrary points \[ (x_1,y_1),(x_2,y_2)\in U_0\times V_0. \] Since \(U_0\) is path-connected, choose a path \[ \alpha:[0,1]\to U_0 \] from \(x_1\) to \(x_2\). Since \(V_0\) is path-connected, choose a path \[ \beta:[0,1]\to V_0 \] from \(y_1\) to \(y_2\). Then \[ \gamma(t)=(\alpha(t),\beta(t)) \] is continuous by the defining property of the product topology, lies in \(U_0\times V_0\), and joins \((x_1,y_1)\) to \((x_2,y_2)\). Thus \(U_0\times V_0\) is path-connected.
4. The product \(X\times Y\) is locally path-connected.
By 2, \(U_0\times V_0\) is open and contains \((x_0,y_0)\). By <1>1–<1>2, \[ U_0\times V_0\subseteq U\times V\subseteq W, \] and by 3 it is path-connected. Thus every open neighborhood of every point of \(X\times Y\) contains a path-connected open neighborhood of that point, which is exactly local path-connectedness.