Here is another theorem whose proof uses Zorn’s lemma. Recall that if \(A\) is a space and if \(x, y \in A\), we say that \(x\) and \(y\) belong to the same quasicomponent of \(A\) if there is no separation \(A = C \cup D\) of \(A\) into two disjoint sets open in \(A\) such that \(x \in C\) and \(y \in D\).
Theorem. Let \(X\) be a compact Hausdorff space. Then \(x\) and \(y\) belong to the same quasicomponent of \(X\) if and only if they belong to the same component of \(X\).
Let \(\mathcal{A}\) be the collection of all closed subspaces \(A\) of \(X\) such that \(x\) and \(y\) lie in the same quasicomponent of \(A\). Let \(\mathcal{B}\) be a subcollection of \(\mathcal{A}\) that is simply ordered by proper inclusion. Show that the intersection of the elements of \(\mathcal{B}\) belongs to \(\mathcal{A}\). [Hint: Compare Exercise 11 of §26.]
Show \(\mathcal{A}\) has a minimal element \(D\).
Show \(D\) is connected.
Solution
Goal: Prove that in a compact Hausdorff space \(X\), two points \(x, y \in X\) belong to the same quasicomponent if and only if they belong to the same connected component.
1. Part (a): Chains in \(\mathcal{A}\) have lower bounds in \(\mathcal{A}\).Proof:
1.1. Let \(\mathcal{B} \subseteq \mathcal{A}\) be a chain ordered by inclusion, and let \(B_\infty = \bigcap_{B \in \mathcal{B}} B\).
1.2. As an intersection of closed subsets containing \(\{x, y\}\), \(B_\infty\) is a closed (hence compact) subspace of \(X\) containing \(x\) and \(y\).
1.3. Suppose for contradiction that \(x\) and \(y\) do not belong to the same quasicomponent of \(B_\infty\).
1.4. Then there exists a separation \(B_\infty = C \cup D\), where \(C, D\) are disjoint closed subsets of \(B_\infty\) with \(x \in C\) and \(y \in D\).
1.5. Since \(B_\infty\) is closed in the normal space \(X\), \(C\) and \(D\) are disjoint closed subsets of \(X\). By normality, choose disjoint open sets \(U, V \subset X\) with \(C \subseteq U\) and \(D \subseteq V\).
1.6. Then \(B_\infty \subseteq U \cup V\), so \(\bigcap_{B \in \mathcal{B}} (B \setminus (U \cup V)) = \varnothing\).
1.7. By the Finite Intersection Property of compact closed sets in \(X\), there exists \(B_0 \in \mathcal{B}\) such that \(B_0 \subseteq U \cup V\).
1.8. Then \(B_0 \cap U\) and \(B_0 \cap V\) form a separation of \(B_0\) into disjoint open sets with \(x \in B_0 \cap U\) and \(y \in B_0 \cap V\), contradicting \(B_0 \in \mathcal{A}\).
1.9. Hence \(x\) and \(y\) lie in the same quasicomponent of \(B_\infty\), so \(B_\infty \in \mathcal{A}\).
2. Part (b): Existence of a minimal element \(D \in \mathcal{A}\).Proof:
2.1. The collection \(\mathcal{A}\) is non-empty because \(X\) is closed and \(x, y\) are in the same quasicomponent of \(X\).
2.2. Order \(\mathcal{A}\) by reverse inclusion (\(A_1 \le A_2 \iff A_1 \supseteq A_2\)).
2.3. By Part (a), every chain in \(\mathcal{A}\) has an upper bound under this ordering (its intersection).
2.4. By Zorn’s Lemma, \(\mathcal{A}\) contains a maximal element with respect to reverse inclusion, which is a minimal element \(D \in \mathcal{A}\) with respect to inclusion.
3. Part (c): Connectedness of \(D\) and equality of components and quasicomponents.Proof:
3.1. Suppose for contradiction that \(D\) is disconnected, so \(D = C_1 \cup D_1\) is a separation of \(D\) into disjoint, non-empty closed sets.
3.2. Since \(x\) and \(y\) are in the same quasicomponent of \(D\), they cannot be separated by this clopen partition; hence both \(x, y \in C_1\) (without loss of generality).
3.3. \(C_1\) is a closed subset of \(D\) (hence of \(X\)). Because \(D_1 \neq \varnothing\), \(C_1 \subsetneq D\) is a strictly smaller closed subset.
3.4. If \(x\) and \(y\) were separated in \(C_1\) by \(C_1 = E \cup F\), then \(D = E \cup (F \cup D_1)\) would be a separation of \(D\) separating \(x\) and \(y\), contradicting \(D \in \mathcal{A}\).
3.5. Thus \(x, y\) lie in the same quasicomponent of \(C_1\), which implies \(C_1 \in \mathcal{A}\).
3.6. This contradicts the minimality of \(D\) in \(\mathcal{A}\).
3.7. Therefore \(D\) is connected.
3.8. Since \(D\) is a connected subspace containing both \(x\) and \(y\), \(x\) and \(y\) belong to the same connected component of \(X\).
4. Conclusion:In any compact Hausdorff space, the quasicomponent of any point coincides with its connected component. Q.E.D.