Show that:
Connected does not imply path connected
Connected and locally path connected does imply path connected
Path connected implies connected
Solution
1. Path-connected spaces are connected.
Fix \(x_0\in X\). For each \(x\in X\), choose a path from \(x_0\) to \(x\). Its image is connected, and all these images contain \(x_0\). Their union is \(X\), so \(X\) is connected.
2. Connected need not imply path-connected.
Consider the closed topologist’s sine curve \[ S=\left\{(x,\sin(1/x)):0<x\le1\right\}\cup(\{0\}\times[-1,1]). \] The graph part is connected as the continuous image of \((0,1]\), and \(S\) is its closure, so \(S\) is connected.
Suppose a path \(\gamma(t)=(x(t),y(t))\) joined a point of the vertical segment to a point of the graph. Let \[ t_0=\sup\{t:x(t)=0\}. \] Then \(t_0<1\), \(x(t_0)=0\), and \(x(t)>0\) for \(t>t_0\) sufficiently near \(t_0\). Since \(x(t)\to0\) as \(t\downarrow t_0\), continuity and the intermediate value theorem force \(x(t)\) arbitrarily near \(t_0\) to take values \[ \frac1{\pi/2+2\pi n} \quad\text{and}\quad \frac1{3\pi/2+2\pi n}, \] where on the graph \(y=1\) and \(y=-1\), respectively. Hence \(y(t)\) takes both values arbitrarily close to \(t_0\), contradicting continuity at \(t_0\). Thus \(S\) is not path-connected.
3. Connected and locally path-connected implies path-connected.
Fix \(x_0\in X\) and let \(P\) be its path component. Local path-connectedness implies every path component is open: a path-connected open neighborhood of any point lies in that point’s path component. Hence \(P\) is open, and its complement, being the union of the other open path components, is open as well. Thus \(P\) is clopen. Since \(X\) is connected and \(P\ne\varnothing\), we have \(P=X\).