Give an example of a topological space that is connected but not path-connected. Make sure you prove that your example satisfies the property.
Solution
1. Let\[ G= \left\{ \left(x,\sin\frac1x\right):0<x\le1 \right\} \subseteq\mathbb R^2 \]and define\[ X=G\cup\{(0,0)\}. \]Then \(G\) is connected.
The interval \((0,1]\) is connected, and the map \[ (0,1]\longrightarrow\mathbb R^2, \qquad x\longmapsto\left(x,\sin\frac1x\right) \] is continuous. Since a continuous image of a connected space is connected, \(G\) is connected.
2. The point \((0,0)\) lies in \(\overline G\), and therefore \(X\) is connected.
For every positive integer \(k\), \[ \left(\frac1{k\pi},0\right)\in G, \] and these points converge to \((0,0)\). Hence \((0,0)\in\overline G\).
To verify connectedness directly, suppose that \[ X=U\cup V \] were a separation of \(X\) into disjoint nonempty open subsets in the subspace topology. Because \(G\) is connected, it must lie entirely in one member of the separation; after relabeling, say \(G\subseteq U\). Then the remaining point \((0,0)\) lies in \(V\). But \(V\) is an open neighborhood of \((0,0)\) in \(X\), while \((0,0)\in\overline G\), so \(V\cap G\ne\varnothing\). This contradicts \(G\subseteq U\) and \(U\cap V=\varnothing\). Thus \(X\) is connected.
3. There is no path in \(X\) from \((0,0)\) to\[ q=\left(\frac1\pi,0\right)\in G. \]
Suppose, for contradiction, that \[ \gamma(t)=(x(t),y(t)):[0,1]\to X \] is continuous with \[ \gamma(0)=(0,0), \qquad \gamma(1)=q. \] The set \[ Z=\{t\in[0,1]:x(t)=0\} \] is nonempty and closed, hence compact. Let \[ t_0=\max Z. \] Since \(x(1)=1/\pi>0\), one has \(t_0<1\). By maximality of \(t_0\), \[ x(t)>0 \qquad(t_0<t\le1). \] Every point of \(X\) with positive first coordinate lies in \(G\), so \[ y(t)=\sin\frac1{x(t)} \qquad(t_0<t\le1). \]
Choose any sufficiently large integer \(m\) for which \(t_0+1/m\le1\). Because \(x(t_0+1/m)>0\), choose numbers \[ 0<a_m,b_m<x(t_0+1/m) \] so small that \[ \sin\frac1{a_m}=1, \qquad \sin\frac1{b_m}=-1. \] Such choices exist because the values \(1\) and \(-1\) occur for reciprocals of arbitrarily large numbers of the forms \[ \frac\pi2+2\pi k, \qquad \frac{3\pi}2+2\pi k. \] Since \[ x(t_0)=0 \] and \(x\) is continuous, the intermediate value theorem gives \[ r_m,s_m\in(t_0,t_0+1/m) \] with \[ x(r_m)=a_m, \qquad x(s_m)=b_m. \] Therefore \[ y(r_m)=1, \qquad y(s_m)=-1. \] But \[ r_m\to t_0, \qquad s_m\to t_0, \] so continuity of \(y\) at \(t_0\) would force both sequences \(y(r_m)\) and \(y(s_m)\) to converge to the same value \(y(t_0)\). This is impossible. Hence no such path exists.
4. Thus \(X\) is connected but not path-connected.
Connectedness is 2, and 3 exhibits two points of \(X\) that cannot be joined by a path.