Let \(X\) be a space. Let us define \(x \sim y\) if there is no separation \(X = A \cup B\) of \(X\) into disjoint open sets such that \(x \in A\) and \(y \in B\).
Show this relation is an equivalence relation. The equivalence classes are called the quasicomponents of \(X\).
Show that each component of \(X\) lies in a quasicomponent of \(X\), and that the components and quasicomponents of \(X\) are the same if \(X\) is locally connected.
Let \(K\) denote the set \(\ts{1/n \mid n \in \mathbb{Z}_+}\) and let \(-K\) denote the set \(\ts{-1/n \mid n \in \mathbb{Z}_+}\). Determine the components, path components, and quasicomponents of the following subspaces of \(\mathbb{R}^2\):
\[ A = (K \times [0, 1]) \cup \ts{0 \times 0} \cup \ts{0 \times 1}, \]
\[ B = A \cup ([0, 1] \times \ts{0}), \]
\[ C = (K \times [0, 1]) \cup (-K \times [-1, 0]) \cup ([0, 1] \times -K) \cup ([-1, 0] \times K). \]
Solution
A separation \(X=A\cup B\) with \(x\in A\) and \(y\in B\) is the same as a set \(A\), open and closed in \(X\), containing \(x\) and not \(y\). So \(x\sim y\) if and only if \(y\) lies in every clopen set containing \(x\). For \(n\in\mathbb Z_+\) write \(L_n=\{1/n\}\times[0,1]\).
1. (a) \(\sim\) is an equivalence relation.
It is reflexive and symmetric because the condition “no clopen set contains one point and not the other” is symmetric and holds for \(x=y\). If \(x\sim y\), \(y\sim z\), and a clopen set \(U\) contains \(x\) but not \(z\), then \(y\in U\) because \(x\sim y\), and then \(z\in U\) because \(y\sim z\), a contradiction.
2. (b) Each component lies in a quasicomponent, and if \(X\) is locally connected the two coincide.
Let \(C(x)\) be the component of \(x\) and \(U\) a clopen set containing \(x\). Then \(C(x)\cap U\) and \(C(x)-U\) are relatively open and disjoint with union \(C(x)\), so \(C(x)-U=\varnothing\) by connectedness. Hence \(C(x)\) lies in every clopen set containing \(x\), that is, in the quasicomponent of \(x\). If \(X\) is locally connected, components are open; then \(C(x)\) is open and its complement, a union of components, is open, so \(C(x)\) is itself a clopen set containing \(x\) and the quasicomponent of \(x\) lies in \(C(x)\).
3. In \(A\), the components and path components are the segments \(L_n\) and the points \(\{(0,0)\}\) and \(\{(0,1)\}\); the quasicomponents are the \(L_n\) and \(\{(0,0),(0,1)\}\).
Each \(L_n\) is path connected, closed, and open in \(A\), since \(L_n=A\cap(V_n\times\mathbb R)\) for an open interval \(V_n\) containing \(1/n\) but neither \(0\) nor any other \(1/k\); so \(L_n\) is a component, a path component, and a quasicomponent. A clopen \(U\ni(0,0)\) contains \((1/n,0)\) for large \(n\), hence \(L_n\), hence \((1/n,1)\) for large \(n\), hence the limit \((0,1)\) because \(U\) is closed; so \((0,0)\sim(0,1)\), and the sets \(L_n\) separate these two points from every other point. By step 2 the components lie in the quasicomponents, and \(\{(0,0),(0,1)\}\) is not connected, so \(\{(0,0)\}\) and \(\{(0,1)\}\) are components and path components.
4. In \(B\), the path components are the comb \(P=(K\times[0,1])\cup([0,1]\times\{0\})\) and \(\{(0,1)\}\), and \(B\) is the single component and quasicomponent.
\(P\) is path connected: each point \((x,y)\) of \(P\) is joined to \((x,0)\) within \(P\) and then along \([0,1]\times\{0\}\). Since \((0,1)=\lim(1/n,1)\) lies in \(\overline P\) and \(P\subseteq B\subseteq\overline P\), \(B\) is connected; by step 2 it is also a single quasicomponent. Let \(\gamma\colon[0,1]\to B\) be a path with \(\gamma(0)=(0,1)\), and let \(T=\gamma^{-1}(\{(0,1)\})\), which is closed. If \(t\in T\), some interval \(J\) about \(t\) has \(\gamma(J)\) in the box \(W=(-\frac12,\frac12)\times(\frac12,\frac32)\); the set \(B\cap W\) is \(\{(0,1)\}\) together with the disjoint segments \(\{1/n\}\times(\frac12,1]\) for \(n\ge3\), each open and closed in the intersection, so the connected set \(\gamma(J)\) containing \((0,1)\) is \(\{(0,1)\}\). Thus \(T\) is open, so \(T=[0,1]\) and \(\gamma\) is constant.
5. In \(C\), the path components are the individual segments, and \(C\) is the single component and quasicomponent.
Let \(U\) be clopen in \(C\) and nonempty; it contains a point of some segment, hence that whole segment, since segments are connected. If \(U\supseteq L_n\), then \(U\) contains \((1/n,-1/m)\) for large \(m\), hence \([0,1]\times\{-1/m\}\) for large \(m\); letting \(m\to\infty\) in \((1/k,-1/m)\in U\) gives \((1/k,0)\in U\), so \(L_k\subseteq U\) for every \(k\). In the same way, \(L_k\subseteq U\) for all \(k\) gives \((0,1/j)=\lim_k(1/k,1/j)\in U\) and so \([-1,0]\times\{1/j\}\subseteq U\) for all \(j\); this gives \((-1/k,0)\in U\) and so \(\{-1/k\}\times[-1,0]\subseteq U\) for all \(k\); and this gives \((0,-1/m)\in U\) and so \([0,1]\times\{-1/m\}\subseteq U\) for all \(m\). Starting from a segment of any of the four families, the same chain of limits shows \(U=C\). So \(C\) has no separation: it is connected and is a single quasicomponent. The segments are pairwise disjoint and closed in \(C\), and there are countably many; for a path \(\gamma\colon[0,1]\to C\), the preimages of the segments partition \([0,1]\) into countably many disjoint closed sets, so by Sierpiński’s theorem only one of them is nonempty. Hence each path lies in one segment, and the segments, being path connected, are the path components.
6. Q.E.D.
Steps 1 and 2 prove (a) and (b), and steps 3, 4, and 5 answer (c).