J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 32: Normal Spaces

Theorem 32.1

Every second countable regular space is normal.

Proof

• Let \(X\) be a second countable regular space.
• Pick a countable basis \(\mathcal{B}\).
• Let \(A\) and \(B\) be disjoint closed sets in \(X\).
• Pick a countable covering \(\{U_1, U_2, \ldots\}\) of \(A\) by open sets whose closures are disjoint from \(B\).
• Let \(\{U_1, U_2, \ldots\} = \{U \in \mathcal{B} \mid \overline{U} \cap B = \emptyset\}\). Prove: this covers \(A\).
• Let \(x \in A\). Prove: there exists \(U \in \mathcal{B}\) such that \(x \in U\) and \(\overline{U} \cap B = \emptyset\).
• Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(B \subseteq V\).
• By regularity.
• Pick \(U' \in \mathcal{B}\) such that \(x \in U' \subseteq U\).
• \(x \in U'\) and \(\overline{U'} \cap B \subseteq \overline{U} \cap B \subseteq (X - V) \cap B = \emptyset\)
• Pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets whose closures are disjoint form \(A\).
• Similar.
• For \(n \in \mathbb{Z}_+\), let \(U_n' = U_n - \bigcup_{i=1}^n \overline{V_i}\) and \(V_n' = V_n - \bigcup_{i=1}^n \overline{U_i}\).
• Each \(U_n'\) and \(V_n'\) is open.
• Since \(U_n\) is open and each \(\overline{V_i}\) is closed.
• \(\{U_1', U_2', \ldots\}\) covers \(A\).
• Let \(x \in A\)
• Pick \(n\) such that \(x \in U_n\)
• Each \(\overline{V_i}\) is disjoint from \(A\).
• \(x \in U_n'\)
• \(\{V_1', V_2', \ldots\}\) covers \(B\).
• Similar.
• Let \(U = \bigcup_n U_n'\) and \(V = \bigcup_n V_n'\).
• \(U\) and \(V\) are disjoint open sets.
• Assume for a contradiction \(x \in U \cap V\)
• Pick \(m\) and \(n\) such that \(x \in U_m' \cap V_n'\).
• Assume w.l.o.g. \(m \leq n\).
• \(x \in V_n'\) and \(x \in \overline{U_m}\).
• This is a contradiction.
• \(A \subseteq U\) and \(B \subseteq V\).
• \(\Box\)

Theorem 32.2

Every metrizable space is normal.

Proof

• Let \(X\) be a metric space.
• Let \(A\) and \(B\) be disjoint closed sets in \(X\).
• For \(a \in A\), choose \(\epsilon_a > 0\) such that \(B(a, \epsilon_a) \cap B = \emptyset\).
• \( \langle 1 \rangle 1 \) For \(b \in B\), choose \(\delta_b > 0\) such that \(B(b, \delta_b) \cap A = \emptyset\).
• Let \(U = \bigcup_{a \in A} B(a, \epsilon_a / 2)\)
• Let \(V = \bigcup_{b \in B) B(b, \delta_b / 2)\)
• \(U\) and \(V\) are open sets with \(A \subseteq U\) and \(B \subseteq V\).
• \(U \cap V = \emptyset\)
• Assume for a contradiction \(z \in U \cap V\).
• Pick \(a \in A\) and \(b \in B\) such that \(d(a,z) < \epsilon_a/2\) and \(d(b,z) < \delta_b/2\).
• Assume w.l.o.g. \(\epsilon_a \leq \delta_b\).
• \(d(a,b) < \delta_b\)
• \(a \in B(b,\delta_b)\)
• This contradicts \(\langle 1 \rangle 1 \).
• \(\Box\)

Theorem 32.3

Every compact Hausdorff space is normal.

Proof

• Let \(X\) be a compact Hausdorff space.
• Let \(A\) and \(B\) be disjoint closed sets in \(X\).
• \(A\) and \(B\) are compact.
• For all \(a \in A\), there exist disjoint open sets \(U\) and \(V\) such that \(a \in U\) and \(B \subseteq V\).
• Lemma 26.4
• \(\{U \text{ open} \mid \exists V \text{ open}. U \cap V = \emptyset, B \subseteq V \}\) is a set of open sets that covers \(A\).
• Pick a finite subset \(\{U_1, \ldots, U_n\}\) that covers \(A\).
• For \(i = 1, \ldots, n\), pick \(V_i\) disjoint from \(U_i\) such that \(B \subseteq V_i\)
• Let \(U = U_1 \cup \cdots \cup U_n\) and \(V = V_1 \cap \cdots \cap V_n\).
• \(U\) and \(V\) are disjoint.
• \(A \subseteq U\) and \(B \subseteq V\)
• \(\Box\)

Exercise 6

A space \(X\) is said to be completely normal if every subspace of \(X\) is normal. Show that \(X\) is completely normal if and only if for every pair \(A\), \(B\) of separated sets in \(X\) (that is, sets such that \(\overline{A} \cap B = \emptyset\) and \(A \cap \overline{B} = \emptyset\)), there exist disjoint open sets containing them.

Solution

• If \(X\) is completely normal then, for every pair \(A\), \(B\) of separated sets, there exist disjoint open sets containing them.
• Assume \(X\) is completely normal.
• Let \(A\) and \(B\) be separated sets.
• \(\overline{A} - \overline{B}\) and \(\overline{B} - \overline{A}\) are disjoint closed sets in \(X - (\overline{A} \cap \overline{B})\).
• Pick disjoint open sets \(U\) and \(V\) in \(X - (\overline{A} \cap \overline{B})\) such that \(\overline{A} - \overline{B} \subseteq U\) and \(\overline{B} - \overline{A} \subseteq V\).
• \(U\) and \(V\) are open in \(X\).
• If every pair of separated sets have disjoint open sets containing them, then \(X\) is completely normal.
• Assume every pair of separated sets have disjoint open sets containing them.
• Let \(Y \subseteq X\).
• Let \(A\) and \(B\) be disjoint closed sets in \(Y\).
• \(A\) and \(B\) are separated sets in \(X\).
• Pick closed sets \(A'\) and \(B'\) in \(X\) such that \(A = A' \cap Y\) and \(B = B' \cap Y\).
• \(\overline{A} \cap B = \emptyset\)
• \(\overline{A} \cap B \subseteq A' \cap B = A' \cap B' \cap Y = A \cap B = \emptyset\)
• \(A \cap \overline{B} = \emptyset\)
• Similar.
• Pick disjoint open sets \(U\) and \(V\) including \(A\) and \(B\) respectively.
• \(U \cap Y\) and \(V \cap Y\) are disjoint open sets in \(Y\) including \(A\) and \(B\) respectively.
• \(\Box\)

Theorem 32.4

Every linearly ordered set is completely normal in the order topology.

Proof

• Let \(X\) be a linearly ordered set.
• The union of any set of convex sets with nonempty intersection is convex.
• Any subset of \(X\) can be written as a union of pairwise disjoint, nonempty, maximal convex sets.
• Let the set of convex components of a subset \(S\) of \(X\) be the set of pairwise disjoint, nonempty, maximal convex sets whose union is \(S\).
• Let \(A\) and \(B\) be separated sets in \(X\).
• Let \(A^* = \bigcup \{[a,b] \mid a,b \in A, [a,b] \cap \overline{B} = \emptyset\}\).
• Let \(B^* = \bigcup \{[a,b] \mid a,b \in B, [a,b] \cap \overline{A} = \emptyset\}\).
• \(A \subseteq A^*\)
• For \(a \in A\) we have \([a,a] \cap \overline{B} = \emptyset\).
• \(B \subseteq B^*\)
• Similar.
• \( \langle 1 \rangle 1\) \(A^* \cap B^* = \emptyset\)
• Assume for a contradiction \(p \in A^* \cap B^*\).
• Pick \(a,b \in A\) such that \(p \in [a,b]\) and \([a,b] \cap \overline{B} = \emptyset\)
• Pick \(c,d \in B\) such that \(p \in [c,d]\) and \([c,d] \cap \overline{A} = \emptyset\)
• Assume w.l.o.g. \(a \leq c\)
• \(c \in [a,b]\)
• This is a contradiction.
• \(A^*\) and \(B^*\) are separated.
• \(\overline{A^*} \cap B^* = \emptyset\)
• \(\overline{A^*} \subseteq A^* \cup \overline{A}\)
• Let \(p \notin A^* \cup \overline{A}\). Prove: \(p \notin \overline{A^*}\).
• Pick an open interval \((s,t)\) containing \(p\) that does not intersect \(A\). Prove: \((s,t)\) does not intersect \(A^*\).
• Assume for a contradiction \(x \in (s,t) \cap A^*\).
• Pick \(a,b \in A\) such that \(x \in [a,b]\) and \([a,b] \cap \overline{B} = \emptyset\).
• \((s,t) \subseteq (a,b) \subseteq A^*\)
• \(p \in A^*\)
• This is a contradiction.
• \(\overline{A^*} \cap B^* = \emptyset\)
\[ \begin{align} \overline{A^*} \cap B^* & \subseteq (A^* \cup \overline{A}) \cap B^* \\ & = (A^* \cap B^*) \cup (\overline{A} \cap B^*) \\ & = \emptyset & (langle 1 \rangle 1) \end{align} \]
• \(A^* \cap \overline{B^*} = \emptyset\)
• Similar.
• Let \(C_A\) be the set of convex components of \(A^*\).
• Let \(C_B\) be the set of convex components of \(B^*\).
• Let \(C_C\) be the set of convex components of \(X - (A^* \cup B^*)\).
• Let \(M = C_A \cup C_B \cup C_C\).
• \(M\) inherits a linear order from \(X\).
• For \(c \in C_A \cup C_B\), let \(u_c\) be the set of strict upper bounds for \(c\), and \(l_c\) the set of strict lower bounds.
• For \(c \in C_A \cup C_B\), if \(c \cap \overline{u_c} \neq \emptyset\) then \(c\) has a successor \(c^+ \in M\) that is an element of \(C_C\).
• Let \(c \in C_A \cup C_B\).
• Assume w.l.o.g. \(c \in C_A\).
• Assume \(c \cap \overline{u_c} \neq \emptyset\).
• Let \(c \cap \overline{u_c} = \{p\}\).
• Assume for a contradiction \(p,q \in c \cap \overline{u_c}\) with \(p < q\).
• \((- \infty, q)\) is a neighbourhood of \(p\)
• Pick \(u \in u_c \cap (-\infty, q)\)
• \(q < u\)
• Because \(q \in c\) and \(u \in u_c\).
• \(u < q\)
• This is a contradiction.
• \(p \notin \overline{B^*}\)
• Because \(p \in A^*\) and \(A^*\) and \(B^*\) are separated.
• \( \langle 2 \rangle 1 \) Pick an open interval \((x,y)\) containing \(p\) that does not intersect \(\overline{B^*}\).
• \((x,y) \cap u_c \neq \emptyset\)
• It is a neighbourhood of \(p \in \overline{u_c}\).
• \((p,y) \neq \emptyset\)
• Pick \(v \in (x,y) \cap u_c\)
• \(p < v\)
• Because \(p \in c\) and \(v \in u_c\)
• \(v \in (p,y)\)
• \( \langle 2 \rangle 2 \) \((p,y) \cap B^* = \emptyset\)
• From \(\langle 2 \rangle 1 \).
• \((p,y) \cap A^* = \emptyset\)
• Assume for a contradiction \(q \in (p,y) \cap A^*\)
• \((-\infty, q)\) is a neighbourhood of \(p\)
• Pick \(u \in u_c \cap (-\infty, q)\)
• \(u \notin c\)
• Pick \(t \in [p,u]\) such that \(t \notin A^*\)
• \([p,u]\) intersects \(B^*\)
• \((p,u]\) intersects \(B^*\)
• This contradicts \langle 2 \rangle 2.
• Let \(c^+ \in C_C\) be the convex component that includes \((p,y)\)
• \(c^+\) is the successor of \(c\) in \(M\).
• For \(c \in C_A \cup C_B\), if \(c \cap \overline{l_c} \neq \emptyset\) then \(c\) has a predecessor \(c^- \in M\) that is an element of \(C_C\).
• Similar.
• For \(c \in C_C\), pick a point \(k_c \in c\).
• For \(c \in C_A \cup C_B\), if \(c \cap \overline{u_c} \neq \emptyset\) then let \(c \cap \overline{u_c} = \{p\}\) and let \(I_c = [p,k_{c^+})\); otherwise let \(I_c = \emptyset\).
• For \(c \in C_A \cup C_B\), if \(c \cap \overline{l_c} \neq \emptyset\) then pick \(c \cap \overline{l_c} = \{p\}\) and let \(J_c = (k_{c^-},p]\); otherwise let \(J_c = \emptyset\).
• For \(c \in C_A \cup C_B\), let \(U_c = J_c \cup c \cup I_c\).
• For \(c \in C_A \cup C_B\), we have \(U_c\) is a convex open set including \(c\).
• Let \(U = \bigcup_{c \in C_A} U_c\).
• Let \(V = \bigcup_{c \in C_B} U_c\).
• \(U\) and \(V\) are open sets including \(A^*\) and \(B^*\) respectively.
• \(U \cap V = \emptyset\)
• Assume for a contradiction \(x \in U \cap V\)
• Pick \(c_1 \in C_A\) and \(c_2 \in C_B\) such that \(x \in U_{c_1} \cap U_{c_2}\).
• Assume w.l.o.g. \(x \in J_{c_1} \cap I_{c_2}\)
• Let \(c = c_1^+ = c_2^-\)
• \(x < k_c\) and \(x > k_c\)
• This is a contradiction.
• \(\Box\)

Example 1 and Exercise 9 (A. H. Stone)

If \(J\) is uncountable then \(\mathbb{R}^J\) is not normal. Thus, not every regular space is normal. A subspace of a normal space is not necessarily normal (as \(\mathbb{R}^J \cong (0,1)^J \subseteq [0,1]^J\)). A product of normal spaces is not necessarily normal.

Proof

• Let \(X = \mathbb{Z}_+^J\).
• \(X\) is a closed subspace of \(\mathbb{R}^J\). Prove: \(X\) is not normal.
• For \(x \in X\) and \(B \subseteq^\mathrm{fin} J\), let \(U(x,B) = \{y \in X \mid \forall \alpha \in B. x(\alpha) = y(\alpha)\}\).
• \(\{U(x,B) \mid x \in X, B \subseteq^\mathrm{fin} J\}\) is a basis for \(X\).
• Let \(x \in X\) and \(U\) be a neighbourhood of \(x\).
• Pick \(B \subseteq^\mathrm{fin} J\) and a basic open set \(\prod_\alpha U_\alpha\) where \(U_\alpha = \mathbb{Z}_+\) for all \(\alpha \notin B\) such that \(x \in \prod_\alpha U_\alpha \subseteq U\).
• \(x \in U(x,B) \subseteq U_\alpha \subseteq U\)
• For \(n \in \mathbb{Z}_+\), let \(P_n = \{x \in X \mid x \text{ is injective on } J - x^{-1}(n)\}\).
• \(P_1\) and \(P_2\) are closed and disjoint.
• For all \(n\), \(P_n\) is closed.
• Let \(n \in \mathbb{Z}_+\).
• Let \(x \in X - P_n\).
• Pick \(\alpha, \alpha' \in J - x^{-1}(n)\) such that \(x(\alpha) \neq x(\alpha')\)
• \(U(x,\{\alpha, \alpha'\}) \subseteq X - P_n\)
• \(P_1 \cap P_2 = \emptyset\)
• Let \(x \in P_1 \cap P_2\)
• \(x\) is injective on \(J - x^{-1}(1)\) and \(J - x^{-1}(2)\).
• \(x\) is an injective function \(J \rightarrow \mathbb{Z}_+\).
• This contradicts the fact that \(J\) is uncountable.
• Let \(U\) and \(V\) be open sets including \(P_1\) and \(P_2\) respectively. Prove: \(U \cap V \neq \emptyset\).
• For any sequence \((\alpha_n)\) of elements of \(J\) and strictly increasing sequence \((n_k)\) of positive integers, define \(B((\alpha_n),(n_k),i) \subseteq^\mathrm{fin} J\) and \(x((\alpha_n),(n_k),i) \in X\) by: \[ \begin{align} B((\alpha_n),(n_k),i) & = \{\alpha_1, \alpha_2, \ldots, \alpha_{n_i}\} \\ x((\alpha_n),(n_k),i)(\alpha_j) & = j & (1 \leq j \leq n_{i-1}) \\ B((\alpha_n),(n_k),i)(\alpha) & = 1 & (\text{for all other values of } \alpha) \end{align} \]
• Choose sequences \((\alpha_n)\) and \((n_k)\) such that, for all \(i\), we have \(U(x((\alpha_n),(n_k),i),B((\alpha_n),(n_k),i)) \subseteq U\).
• Assume as induction hypothesis we have chosen sequences \((n_1, \ldots n_k)\) \((k \geq 0)\) and \((\alpha_1, \alpha_2, \ldots, \alpha_{n_k})\) such that, for all \(i \leq k\), we have \(U(x((\alpha_n),(n_k),i),B((\alpha_n),(n_k),i)) \subseteq U\).
• \(x((\alpha_n),(n_k),k+1) \in P_1 \subseteq U\)
• Pick \(B\) nonempty such that \(U(x((\alpha_n),(n_k),k+1),B) \subseteq U\)
• Let \(B = \{\alpha_{n_k + 1}, \alpha_{n_k + 2}, \ldots, \alpha_{n_{k+1}}\}\)
• For all \(i \leq k+1\) we have \(U(x((\alpha_n),(n_k),i),B((\alpha_n),(n_k),i)) \subseteq U\).
• Let \(A = \{\alpha_1, \alpha_2, \ldots\}\).
• Define \(y \in X\) by \(y(\alpha_j) = j\) for \(\alpha_j \in A\), and \(y(\alpha) = 2\) for all other \(\alpha\).
• Pick \(B\) such that \(U(y,B) \subseteq V\).
• Pick \(i\) such that \(B \cap A \subseteq B((\alpha_n),(n_k),i)\).
• \(U(x((\alpha_n),(n_k),i+1),B((\alpha_n),(n_k),i+1)) \cap U(y,B) \neq \emptyset\)
• \(y\) is a member of both.
• \(U \cap V \neq \emptyset\)
• \(\Box\)

Example 2 (J. Dieudonné and A. P. Morse)

The product space \(\Omega \times (\Omega + 1)\) is not normal. Thus, the product of two normal spaces is not necessarily normal.

• Let \(\Delta = \{(x,x) \mid x \in \Omega + 1\}\).
• \(\Delta\) is closed in \((\Omega + 1)^2\).
• Let \(A = \Delta \cap (\Omega \times (\Omega + 1)) = \Delta - \{(\Omega,\Omega)\}\).
• \(A\) is closed in \(\Omega \times (\Omega + 1)\).
• Let \(B = \Omega \times \{\Omega\}\).
• \(B\) is closed in \(\Omega \times (\Omega + 1)\).
• \(A \cap B = \emptyset\)
• Assume for a contradiction \(U\) and \(V\) are disjoint open sets in \(\Omega \times (\Omega + 1)\) including \(A\) and \(B\) respectively.
• For all \(x < \Omega\), let \(\beta(x)\) be the least point such that \(x < \beta(x) < \Omega\) and \((x,\beta(x)) \notin U\).
• Let \(x < \Omega\). Prove: there exists \(\beta\) such that \(x < \beta < \Omega\) and \((x,\beta) \notin U\).
• \((x,\Omega) \in B \subseteq V\)
• Pick \(\beta < \Omega\) such that \(\{x\} \times (\beta, \Omega] \subseteq V\)
• \((x, \beta) \notin U\)
• Define the sequence \((x_n)\) in \(\Omega\) by \(x_1 = 0\) and \(x_{n+1} = \beta(x_n)\) for all \(n\).
• \(x_1 < x_2 < \cdots\)
• Let \(b = \sup_n x_n\)
• \(x_n \rightarrow b\) as \(n \rightarrow \infty\).
• \(\beta(x_n) \rightarrow b\) as \(n \rightarrow \infty\).
• \((x_n, \beta(x_n)) \rightarrow (b,b)\) as \(n \rightarrow \infty\).
• \((b,b) \in A\)
• \((b,b) \in U\)
• None of the points \((x_n, \beta(x_n))\) is in \(U\).
• This is a contradiction.
• \(\Box\)

Exercises

Exercise 1

Show that a closed subspace of a normal space is normal.

Solution

• Let \(X\) be a normal space and \(A\) be closed in \(X\).
• Let \(B\) and \(C\) be disjoint closed sets in \(A\).
• \(B\) and \(C\) are closed in \(X\).
• Pick disjoint open sets \(U\) and \(V\) in \(X\) that include \(B\) and \(C\).
• \(U \cap A\) and \(V \cap A\) are disjoint open sets in \(A\) that include \(B\) and \(C\).
• \(\Box\)

Exercise 2

Show that if \(\prod_\alpha X_\alpha\) is Hausdorff, or regular, or normal, then so is \(X_\alpha\). (Assume that each \(X_\alpha\) is nonempty.)

Solution

Since each \(X_\alpha\) is homeomorphic to a closed subspace of \(\prod_\alpha X_\alpha\).

Exercise 3

Show that every locally compact Hausdorff space is regular.

Solution

If \(X\) is a locally compact Hausdorff space then it is a subspace of a compact Hausdorff space which is regular, and so \(X\) is regular.

Exercise 4

Show that every regular Lindelöf space is normal.

Solution

• Let \(X\) be a regular Linedlöf space.
• Let \(A\) and \(B\) be disjoint closed sets in \(X\).
• For \(a \in A\), there exist disjoint neighbourhoods of \(a\) and \(B\).
• \(\{U \text{ open in } \overline{U} \cap B = \emptyset \}\) is a set of open sets that covers \(A\).
• Pick a countable subcovering \(\{U_1, U_2, \ldots\}\).
• Similarly, pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets such that, for all \(i\), we have \(\overline{V_i} \cap A = \emptyset\).
• For \(n \in \mathbb{Z}_+\), let \(U_n' = U_n - \bigcup_{i=1}^n \overline{V_i}\) and \(V_n' = V_n - \bigcup_{i=1}^n \overline{U_i}\).
• Let \(U = \bigcup_n U_n'\) and \(V = \bigcup_n V_n'\)
• \(U\) and \(V\) are disjoint neighbourhoods of \(A\) and \(B\).
• \(\Box\)

Exercise 5

Is \(\mathbb{R}^\omega\) normal in the product topology? Is it normal in the uniform topology?

Solution

The answer to both questions is yes because both spaces are metrizable.

We will return to the fact that if the Continuum Hypothesis is true then \(\mathbb{R}^\omega\) is normal in the box topology after we have covered paracompactness.

Exercise 7

Which of the following spaces are completely normal? Justify your answers.

(a)

A subspace of a completely normal space.

Solution

Yes. Let \(X\) be a completely normal space and \(Y\) a subspace of \(X\). Let \(Z\) be any subspace of \(Y\). Then \(Z\) is a subspace of \(X\), hence normal.

(b)

The product of two completely normal spaces.

Solution

Not necessarily. \(\Omega\) and \(\Omega + 1\) are completely normal, being order topologies, but their product is not normal.

(c)

A well-ordered set in the order topology.

Solution

Yes - see above.

(d)

A metrizable space.

Solution

Yes. Any subspace of a metrizable space is metrizable, hence normal.

(e)

A compact Hausdorff space.

Solution

Not necessarily. For \(J\) uncountable, \(\mathbb{R}^J\) is a subspace of the compact Hausdorff space \([0,1]^J\), but it is not normal.

(f)

A second countable regular space.

Solution

Yes. Any subspace is second countable and regular, hence normal.

(g)

The space \(\mathbb{R}_l\).

Solution

Yes.

• Let \(Y\) be any subspace of \(\mathbb{R}_l\).
• Let \(A\) and \(B\) be disjoint closed sets in \(Y\).
• Pick closed sets \(A'\) and \(B'\) in \(\mathbb{R}_l\) such that \(A = A' \cap Y\) and \(B = B' \cap Y\).
• For \(a \in A\), pick \(x_a > a\) such that \([a,x_a) \subseteq X - B'\).
• For \(b \in B\), pick \(y_b > b\) such that \([b,y_b) \subseteq X - A'\).
• Let \(U = \bigcup_{a \in A} [a,x_a)\) and \(V = \bigcup_{b \in B} [b,y_b)\).
• \(U \cap Y\) and \(B \cap Y\) are disjoint neighbourhoods of \(A\) and \(B\).
• \(\Box\)

Exercise 10

Is every topological group normal?

Solution

No - \(\mathbb{R}^J\) for \(J\) uncountable is not normal.