For a topological space \(X\) and \(y\in X\), the path component \(P_y\) of \(X\) containing \(y\) is the largest path-connected subset with \(y\in P_y\subseteq X\).
Show that this concept is well-defined (that is, show that every point \(y\) is contained in a largest path-connected subset).
Give an example of a space and point \(y\in X\) so that \(P_y\) is neither an open nor a closed subset of \(X\).
Solution
1. For \(y\in X\), define\[ P_y = \{x\in X:\text{there is a path in }X\text{ from }y\text{ to }x\}. \]Then \(P_y\) is path-connected.
Certainly \(y\in P_y\) by the constant path. Let \(x,z\in P_y\). Choose paths \[ \alpha:[0,1]\to X, \qquad \beta:[0,1]\to X \] with \[ \alpha(0)=y,\quad \alpha(1)=x, \qquad \beta(0)=y,\quad \beta(1)=z. \] Reverse \(\alpha\) and concatenate it with \(\beta\). This gives a path from \(x\) to \(z\) whose image lies in \(P_y\), since every point on either path is itself reachable from \(y\) by an initial segment of one of these paths. Thus any two points of \(P_y\) can be joined by a path in \(P_y\).
2. The set \(P_y\) is the largest path-connected subset of \(X\) containing \(y\).
Let \(A\subseteq X\) be path-connected and suppose \(y\in A\). For every \(x\in A\), path-connectedness of \(A\) gives a path in \(A\), hence in \(X\), from \(y\) to \(x\). Therefore \(x\in P_y\) and \[ A\subseteq P_y. \] By 1, \(P_y\) itself is path-connected and contains \(y\), so it is the required largest such subset. This proves part (a).
For part (b), let \[ G = \left\{\left(x,\sin\frac1x\right):0<x\le\frac1\pi\right\}, \qquad V=\{0\}\times[-1,1], \] and let \[ T=G\cup V \] be the closed topologist’s sine curve. Set \[ p=\left(\frac1\pi,0\right)\in G, \qquad q_n=\left(\frac1\pi,\frac1n\right), \] and define the subspace \[ X=T\cup\{q_n:n\ge1\}\subseteq\mathbb R^2. \]
3. Each \(q_n\) is an isolated point of \(X\), and \(q_n\to p\).
The convergence is immediate from the coordinates. The set \(T\) is compact, hence closed in \(\mathbb R^2\), and \(q_n\notin T\) because the only point of \(T\) with first coordinate \(1/\pi\) is \(p\). Thus each \(q_n\) has positive distance from \(T\). Also the sequence \(\{q_m\}\) has no accumulation point other than \(p\), so for fixed \(n\) there is a small Euclidean ball about \(q_n\) containing no other \(q_m\). Intersecting the two small neighborhoods shows \[ \{q_n\} \] is open in the subspace \(X\).
4. The set \(G\) is a path component of \(X\).
The graph \(G\) is path-connected because it is the continuous image of the interval \((0,1/\pi]\) under \[ x\longmapsto\left(x,\sin\frac1x\right). \] For the standard topologist’s sine curve \(T\), there is no path from a point of \(G\) to a point of the vertical segment \(V\); hence \(G\) and \(V\) are the two path components of \(T\).
Adding the isolated points \(q_n\) creates no path from \(G\) to anything outside it. Indeed, the image of a path is connected. If such an image contained \(q_n\) and another point, then \(\{q_n\}\) would be open in the image by 3 and closed in the image because \(X\) is a metric space, contradicting connectedness. Thus a path starting in \(G\) cannot reach any \(q_n\). Nor can it reach \(V\), since any such path cannot use the \(q_n\) and would therefore give a path in \(T\) from \(G\) to \(V\). Hence the path component of \(p\) in \(X\) is exactly \(G\).
5. The path component \(G=P_p\) is not closed in \(X\).
Every point \((0,t)\in V\) is a limit point of \(G\): choose numbers \(x_k\downarrow0\) such that \[ \sin(1/x_k)\to t. \] Hence \[ V\subseteq\overline G^X. \] Since \(V\cap G=\varnothing\), the set \(G\) does not contain all of its limit points and is not closed in \(X\).
6. The path component \(G=P_p\) is not open in \(X\).
By 3, \[ q_n\longrightarrow p\in G, \] while every \(q_n\) lies outside \(G\). Thus every neighborhood of \(p\) in \(X\) meets \(X\setminus G\). Therefore \(p\) is not an interior point of \(G\), so \(G\) is not open.
7. Thus the space \(X\) and the point \(p\) give the example required in part (b): the path component \(P_p=G\) is neither open nor closed.
This is exactly <1>4–<1>6.