Let \(X\) be a space that is the union of subspaces \(S_1, \ldots, S_n\), each of which is homeomorphic to the unit circle. Assume there is a point \(p\) of \(X\) such that \(S_i \cap S_j = \ts{p}\) for \(i \neq j\).
Show that \(X\) is Hausdorff if and only if each space \(S_i\) is closed in \(X\).
Show that \(X\) is Hausdorff if and only if the topology of \(X\) is coherent with the subspaces \(S_i\).
Give an example to show that \(X\) need not be Hausdorff. [Hint: See E-29WP0.]
Solution
- Suppose first that \(X\) is Hausdorff. Each \(S_i\) is compact, being homeomorphic to \(S^1\), so each \(S_i\) is closed in \(X\).
Conversely, suppose each \(S_i\) is closed in \(X\). Then \(S_i\times S_i\) is closed in \(X\times X\), and since \(S_i\) is Hausdorff its diagonal \(\Delta_{S_i}\) is closed in \(S_i\times S_i\), hence in \(X\times X\). Because the cover is finite, \[ \Delta_X=\bigcup_{i=1}^n\Delta_{S_i} \] is closed. Therefore \(X\) is Hausdorff.
- For a finite cover by closed subspaces, the topology is coherent with the cover: a set \(A\subset X\) is closed iff every \(A\cap S_i\) is closed in \(S_i\). Indeed, the forward implication is immediate, while in the reverse direction each \(A\cap S_i\) is closed in \(X\) and \[ A=\bigcup_{i=1}^n(A\cap S_i) \] is a finite union of closed sets. Hence, if \(X\) is Hausdorff, part (a) shows all \(S_i\) are closed and the topology is coherent.
Conversely, assume the topology is coherent with the \(S_i\). Fix \(i\). For every \(j\), \[ S_i\cap S_j= \begin{cases} S_i,&j=i,\\ \{p\},&j\ne i, \end{cases} \] and these are closed in \(S_j\). Coherence therefore implies that \(S_i\) is closed in \(X\). Part (a) now gives that \(X\) is Hausdorff.
- Here is an explicit non-Hausdorff example. Start with two circles \(S_1,S_2\) meeting only at \(p\). Choose points \(q_i\in S_i-\{p\}\) and small open arcs \(A_i\subset S_i-\{p\}\) about \(q_i\), together with homeomorphisms \[ \theta_i:A_i\longrightarrow(-1,1),\qquad \theta_i(q_i)=0. \] Away from \(q_1,q_2\), use the ordinary wedge topology. A basic neighborhood of \(q_1\) is, for \(0<\varepsilon<1\), \[ \theta_1^{-1}((-\varepsilon,\varepsilon)) \;\cup\; \theta_2^{-1}((-\varepsilon,0)\cup(0,\varepsilon)), \] and define basic neighborhoods of \(q_2\) symmetrically. This is exactly the line-with-two-origins construction localized in the two arcs. Its restriction to each \(S_i\) is the usual circle topology, and \(S_1\cap S_2=\{p\}\). But every neighborhood of \(q_1\) meets every neighborhood of \(q_2\), so the resulting \(X\) is not Hausdorff.