Recall that \(\mathbb{R}_K\) denotes the real line in the \(K\)-topology. (See §13.) Let \(Y\) be the quotient space obtained from \(\mathbb{R}_K\) by collapsing the set \(K\) to a point; let \(p: \mathbb{R}_K \to Y\) be the quotient map.
Show that \(Y\) satisfies the \(T_1\) axiom, but is not Hausdorff.
Show that \(p \times p: \mathbb{R}_K \times \mathbb{R}_K \to Y \times Y\) is not a quotient map. [Hint: The diagonal is not closed in \(Y \times Y\), but its inverse image is closed in \(\mathbb{R}_K \times \mathbb{R}_K\).]
Solution
Let \(K=\{1/n:n\in\mathbb Z_+\}\) and let \(q=p(K)\in Y\).
- The space \(\mathbb R_K\) is \(T_1\) because its topology is finer than the usual topology. The set \(K\) is closed in \(\mathbb R_K\): its complement is open since every \(x\notin K\) has a basic neighborhood of the form \((a,b)-K\). Therefore every singleton in \(Y\) has closed inverse image under \(p\): the inverse image of \(\{q\}\) is \(K\), while the inverse image of any other singleton is a singleton of \(\mathbb R_K\). By the quotient topology, singletons in \(Y\) are closed. Hence \(Y\) is \(T_1\).
The space is not Hausdorff. Suppose \(q\) and \(p(0)\) had disjoint open neighborhoods \(U,V\). Then \(p^{-1}(U)\) is an open saturated set containing all of \(K\), and \(p^{-1}(V)\) is an open neighborhood of \(0\). Shrinking inside \(p^{-1}(V)\), choose \[ (-\varepsilon,\varepsilon)-K\subseteq p^{-1}(V). \] Choose \(n\) with \(1/n<\varepsilon\). Since \(1/n\in p^{-1}(U)\) and basic neighborhoods of a point of \(K\) cannot be of the form \((a,b)-K\), openness of \(p^{-1}(U)\) gives an ordinary interval about \(1/n\) contained in it. Such an interval contains points of \((-\varepsilon,\varepsilon)-K\), contradicting disjointness. Thus \(Y\) is not Hausdorff.
- A space is Hausdorff if and only if its diagonal is closed (E-6A0RO). Hence \[ \Delta_Y=\{(y,y):y\in Y\} \] is not closed in \(Y\times Y\). Its inverse image under \(p\times p\) is \[ (p\times p)^{-1}(\Delta_Y)=\Delta_{\mathbb R_K}\cup(K\times K). \] The space \(\mathbb R_K\) is Hausdorff (it is finer than the usual Hausdorff topology), so its diagonal is closed; \(K\) is closed, so \(K\times K\) is closed. Thus this inverse image is closed.
If \(p\times p\) were a quotient map, closedness of a subset of \(Y\times Y\) would be equivalent to closedness of its inverse image. We have found a nonclosed set with closed inverse image, so \(p\times p\) is not quotient.