There is a space that is locally 2-euclidean and satisfies (v) but not (iv) of Exercise 2. It is constructed as follows. Let \(A\) be the following subspace of \(\mathbb{R}^3\):
\[ A = \ts{(x, y, 0) \mid x > 0}. \]
Given \(c\) real, let \(B_c\) be the following subspace of \(\mathbb{R}^3\):
\[ B_c = \ts{(x, y, c) \mid x \leq 0}. \]
Let \(X\) be the set that is the union of \(A\) and all the spaces \(B_c\), for \(c\) real. Topologize \(X\) by taking as a basis all sets of the following three types:
\(U\), where \(U\) is open in \(A\).
\(V\), where \(V\) is open in the subspace of \(B_c\) consisting of points with \(x < 0\).
For each open interval \(I = (a, b)\) of \(\mathbb{R}\), each real number \(c\), and each \(\epsilon > 0\), the set \(A_c(I, \epsilon) \cup B_c(I, \epsilon)\), where
\[ A_c(I, \epsilon) = \ts{(x, y, 0) \mid 0 < x < \epsilon \text{ and } c + ax < y < c + bx}, \]
\[ B_c(I, \epsilon) = \ts{(x, y, c) \mid -\epsilon < x \leq 0 \text{ and } a < y < b}. \]
The space \(X\) is called the “Prüfer manifold.”
Sketch the sets \(A_c(I, \epsilon)\) and \(B_c(I, \epsilon)\).
Show the sets of types (i)-(iii) form a basis for a topology on \(X\).
Show the map \(f_c: \mathbb{R}^2 \to X\) given by
\[ f_c(x, y) = \begin{cases} (x, c + xy, 0) & \text{for } x > 0, \\ (x, y, c) & \text{for } x \leq 0 \end{cases} \]
defines a homeomorphism of \(\mathbb{R}^2\) with the subspace \(A \cup B_c\) of \(X\).
Show that \(A \cup B_c\) is open in \(X\); conclude that \(X\) is 2-euclidean.
Show that \(X\) is Hausdorff.
Show that \(X\) is not normal. [Hint: The subspace
\[ L = \ts{(0, 0, c) \mid c \in \mathbb{R}} \]
of \(X\) is closed and discrete. Compare Example 3 of §31.]
Solution
1. Part (a) & (b): Geometric description and basis verification.Proof:
1.1. Description: In the right half-plane \(A\) (\(z=0, x>0\)), \(A_c(I, \varepsilon)\) is an open wedge radiating from \((0, c, 0)\) bounded by rays of slopes \(a, b\) and width \(x < \varepsilon\). In the sheet \(B_c\) (\(z=c, x \le 0\)), \(B_c(I, \varepsilon)\) is an open rectangular strip \((-\varepsilon, 0] \times (a, b) \times \{c\}\).
1.2. Covering: Every point in \(A\) is covered by (i); every point in \(B_c\) with \(x < 0\) is covered by (ii); every boundary point \((0, y_0, c) \in B_c\) is covered by (iii) with \(I = (y_0 - 1, y_0 + 1)\) and \(\varepsilon = 1\).
1.3. Intersections: Intersections of type (i) with (i) or (iii) are open in \(A\) (type (i)). Intersections of type (ii) with (ii) or (iii) on the same sheet \(B_c\) are open in \(B_c\) (\(x<0\)) (type (ii)). For two type (iii) sets with the same \(c\), their intersection is \(A_c(I_1 \cap I_2, \min(\varepsilon_1, \varepsilon_2)) \cup B_c(I_1 \cap I_2, \min(\varepsilon_1, \varepsilon_2))\) (type (iii)). For distinct \(c_1 \neq c_2\), \(B_{c_1} \cap B_{c_2} = \emptyset\), and the intersection in \(A\) is open in \(A\) (type (i)).
1.4. Thus types (i)-(iii) form a basis.
2. Part (c) & (d): Homeomorphism \(f_c: \mathbb{R}^2 \to A \cup B_c\) and local 2-euclidean structure.Proof:
2.1. \(f_c\) is bijective from \(\mathbb{R}^2\) onto \(A \cup B_c\): on \(x > 0\) it is the bijection onto \(A\) of 2, on \(x \le 0\) it is the bijection \((x, y) \mapsto (x, y, c)\) onto \(B_c\), and the images \(A\) (third coordinate \(0\), \(x > 0\)) and \(B_c\) (\(x \le 0\)) are disjoint.
2.2. On \(x > 0\), \(f_c(x, y) = (x, c + xy, 0)\) is a smooth diffeomorphism onto \(A\) with continuous inverse \((u, v, 0) \mapsto (u, \frac{v-c}{u})\).
2.3. On \(x < 0\), \(f_c(x, y) = (x, y, c)\) is the standard Cartesian identification.
2.4. At \(x = 0\), the image of the basic product neighborhood \((-\varepsilon, \varepsilon) \times (a, b) \subset \mathbb{R}^2\) under \(f_c\) is precisely \(A_c((a, b), \varepsilon) \cup B_c((a, b), \varepsilon)\), which is the type (iii) basis element of \(X\).
2.5. Thus \(f_c\) is a homeomorphism of \(\mathbb{R}^2\) onto \(A \cup B_c\).
2.6. For any \(p \in A \cup B_c\), \(p\) possesses a basis neighborhood contained in \(A \cup B_c\), so \(A \cup B_c\) is open in \(X\). Since \(\{A \cup B_c : c \in \mathbb{R}\}\) is an open cover of \(X\) by Euclidean planes \(\mathbb{R}^2\), \(X\) is locally 2-euclidean.
3. Part (e): \(X\) is Hausdorff.Proof:
3.1. Distinct points in the same sheet \(A \cup B_c\) are separated because \(A \cup B_c \cong \mathbb{R}^2\) is Hausdorff.
3.2. Points \(p \in B_{c_1}\) and \(q \in B_{c_2}\) with \(c_1 \neq c_2\):- If \(x(p) < 0\) or \(x(q) < 0\), disjoint type (ii) neighborhoods separate them.- If \(p = (0, y_1, c_1)\) and \(q = (0, y_2, c_2)\), choose open intervals \(I_1 = (y_1 - 1, y_1 + 1) = (a_1, b_1)\) and \(I_2 = (y_2 - 1, y_2 + 1) = (a_2, b_2)\).- Without loss of generality, assume \(c_1 < c_2\). Choose \(\varepsilon > 0\) such that \(\varepsilon < \frac{c_2 - c_1}{b_1 - a_2}\) (if \(b_1 > a_2\), or any \(\varepsilon > 0\) otherwise).- Then for all \(0 < x < \varepsilon\), \(c_1 + b_1 x < c_2 + a_2 x\), which guarantees that the wedges \(A_{c_1}(I_1, \varepsilon)\) and \(A_{c_2}(I_2, \varepsilon)\) are disjoint.- Since \(B_{c_1} \cap B_{c_2} = \emptyset\), the type (iii) neighborhoods of \(p\) and \(q\) are completely disjoint.
3.3. Thus \(X\) is Hausdorff.
4. Part (f): \(X\) is not normal.Proof:
4.1. The subspace \(L = \{(0, 0, c) : c \in \mathbb{R}\} \subset X\) is closed and discrete, since each basic neighborhood \(A_c((-1, 1), 1) \cup B_c((-1, 1), 1)\) contains only the point \((0, 0, c)\) of \(L\).
4.2. \(L\) is closed: a point of \(A\) has a type (i) neighborhood, a point of \(B_c\) with \(x < 0\) has a type (ii) neighborhood, and a point \((0, y, c)\) with \(y \neq 0\) has a type (iii) neighborhood with \(0 \notin I\); none of these meets \(L\).
4.3. The countable set \(D = (\mathbb{Q}_+ \times \mathbb{Q}) \times \{0\} \subset A\) meets every neighborhood of every point of \(L\), because each such neighborhood contains a type (iii) set, whose part \(A_c(I, \varepsilon)\) is a nonempty open subset of the half-plane \(A\).
4.4. Suppose \(X\) is normal. For each \(S \subseteq L\), the sets \(S\) and \(L \setminus S\) are closed in \(X\) by steps 1 and 2, so there are disjoint open sets \(U_S \supseteq S\) and \(V_S \supseteq L \setminus S\). If \(S \neq T\), say \(p \in S \setminus T\), then \(U_S \cap V_T\) is a neighborhood of \(p\), so it contains a point of \(D\) by step 3; that point lies in \(U_S \cap D\) and not in \(U_T \cap D\). Hence \(S \mapsto U_S \cap D\) is an injective map from the power set of \(L\) to the power set of \(D\).
4.5. The power set of \(L\) has cardinality \(2^{\mathfrak{c}}\) and the power set of the countable set \(D\) has cardinality \(\mathfrak{c} < 2^{\mathfrak{c}}\), which contradicts step
4.4. Therefore \(X\) is not normal. Q.E.D.