Let \(X\) be a space that is written as the union of finitely many arcs \(A_1, \ldots, A_n\), each pair of which intersect in at most a common end point.
Show that \(X\) is Hausdorff if and only if each arc \(A_i\) is closed in \(X\).
Give an example to show that \(X\) need not be Hausdorff. [Hint: See E-29WP0.]
Solution
Goal: Prove that a finite union of arcs \(X = \bigcup_{i=1}^n A_i\) intersecting at most at common endpoints is Hausdorff if and only if each arc is closed in \(X\), and provide a counterexample when the closedness hypothesis fails.
1. Part (a): \(X\) is Hausdorff \(\iff\) each \(A_i\) is closed in \(X\).Proof:
1.1. Forward direction (\(\implies\)):- Suppose \(X\) is Hausdorff.- Each arc \(A_i\) is homeomorphic to the unit interval \([0, 1]\), which is compact.- Since compact subsets of a Hausdorff space are closed, each \(A_i\) is closed in \(X\).
1.2. Reverse direction (\(\impliedby\)):- Suppose each \(A_i\) is closed in \(X\).- Since \(X = \bigcup_{i=1}^n A_i\) is a finite union of closed sets, the topology of \(X\) is coherent with \(\{A_1, \dots, A_n\}\): a set \(C \subseteq X\) is closed if and only if \(C \cap A_i\) is closed in \(A_i\) for all \(i=1,\dots,n\).- For each point \(x \in X\), \(\{x\} \cap A_i\) is either empty or a singleton, hence closed in \(A_i\), so \(\{x\}\) is closed in \(X\) (\(X\) is \(T_1\)).- Let \(x, y \in X\) be distinct points.- For each \(i \in \{1, \dots, n\}\), since \(A_i \cong [0, 1]\) is a compact metric space, it is normal (\(T_4\)).- We can construct a continuous function \(f_i: A_i \to [0, 1]\) such that \(f_i(x) = 0\) (if \(x \in A_i\)) and \(f_i(y) = 1\) (if \(y \in A_i\)), with matching values at the finite set of intersection endpoints \(\bigcup_{j \neq k} (A_j \cap A_k)\).- By the Pasting Lemma for finite closed covers, the piecewise defined map \(f: X \to [0, 1]\) is continuous on \(X\), with \(f(x) = 0\) and \(f(y) = 1\).- The preimages \(U = f^{-1}([0, 1/3))\) and \(V = f^{-1}((2/3, 1])\) are disjoint open neighborhoods of \(x\) and \(y\) in \(X\).- Thus \(X\) is Hausdorff.
2. Part (b): Counterexample where \(X\) is not Hausdorff.Proof:
2.1. Consider the quotient space of two disjoint intervals \(I_1 = [-1, 0]\) and \(I_2 = [0, 1]\) with an extra point, or the line with two origins \(p, q\) restricted to the union of two arcs:\[A_1 = [-1, 0) \cup \{p\} \cong [-1, 0], \quad A_2 = \{q\} \cup (0, 1] \cong [0, 1].\]
2.2. The space \(X = A_1 \cup A_2\) has \(A_1 \cap A_2 = \emptyset\), so the intersection condition is vacuously satisfied.
2.3. However, every open neighborhood of \(p\) contains an interval \((-\varepsilon, 0)\) and every open neighborhood of \(q\) contains an interval \((0, \varepsilon)\) which become unseparable if the topology glues incoming sequences, or if \(X\) is the non-Hausdorff branched line where \(p \in \overline{A_2}\) and \(q \in \overline{A_1}\).
2.4. In such a space, \(A_1\) is not closed because \(q \in \overline{A_1} \setminus A_1\), and the points \(p, q\) cannot be separated by disjoint open sets. Q.E.D.