Is a product of path-connected spaces necessarily path connected?
If \(A \subset X\) and \(A\) is path connected, is \(\overline{A}\) necessarily path connected?
If \(f: X \to Y\) is continuous and \(X\) is path connected, is \(f(X)\) necessarily path connected?
If \(\ts{A_\alpha}\) is a collection of path-connected subspaces of \(X\) and if \(\bigcap A_\alpha \neq \varnothing\), is \(\bigcup A_\alpha\) necessarily path connected?
Solution
1. (a) \(\boxed{\text{Yes}}\): a product \(\prod_{\alpha\in J}X_\alpha\) of path-connected spaces is path connected.
Given \(\mathbf x,\mathbf y\) in the product, choose paths \(\gamma_\alpha\colon[0,1]\to X_\alpha\) from \(x_\alpha\) to \(y_\alpha\). The map \(\gamma(t)=(\gamma_\alpha(t))_\alpha\) is continuous because its coordinates are, and it is a path from \(\mathbf x\) to \(\mathbf y\).
2. (b) \(\boxed{\text{No}}\): the closure of a path-connected set need not be path connected.
The set \(A=\{(x,\sin(1/x)):0<x\le1\}\subseteq\mathbb R^2\) is the image of \((0,1]\) under a continuous map, hence path connected. Its closure \(\overline A=A\cup(\{0\}\times[-1,1])\) is the topologist’s sine curve, which is not path connected: no path in \(\overline A\) joins \((0,0)\) to a point of \(A\).
3. (c) \(\boxed{\text{Yes}}\): if \(f\colon X\to Y\) is continuous and \(X\) is path connected, \(f(X)\) is path connected.
For \(f(x_1),f(x_2)\in f(X)\), choose a path \(\gamma\) in \(X\) from \(x_1\) to \(x_2\); then \(f\circ\gamma\) is a path in \(f(X)\) from \(f(x_1)\) to \(f(x_2)\).
4. (d) \(\boxed{\text{Yes}}\): if the \(A_\alpha\) are path connected and \(p\in\bigcap_\alpha A_\alpha\), then \(\bigcup_\alpha A_\alpha\) is path connected.
For \(x\in A_\beta\) and \(y\in A_\delta\), choose a path in \(A_\beta\) from \(x\) to \(p\) and a path in \(A_\delta\) from \(p\) to \(y\); their concatenation is a path in \(\bigcup_\alpha A_\alpha\) from \(x\) to \(y\).
5. Q.E.D.
Steps 1 through 4 answer (a) through (d).