J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 31: The Separation Axioms

Definition

A topological space \(X\) is regular iff it is \(T_1\) and, for any point \(x \in X\) and closed set \(B\) such that \(x \notin B\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(B \subseteq V\).

A topological space \(X\) is normal iff it is \(T_1\) and, for any disjoint closed sets \(A\) and \(B\), there exist disjoint sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\).

Lemma 31.1

Let \(X\) be a \(T_1\) space.

1. \(X\) is regular if and only if, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\).
2. \(X\) is normal if and only if, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\).

Proof

• If \(X\) is regular then, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\).
• Assume \(X\) is regular.
• Let \(x \in X\) and \(U\) be a neighbourhood of \(x\).
• \(x \notin X - U\)
• Pick disjoint open sets \(V\) and \(W\) such that \(x \in V\) and \(X - U \subseteq W\)
• \(\overline{V} \subseteq U\)
• \(\overline{V} \subseteq X - W \subseteq U\)
• If, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\).
• Assume that, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\).
• Let \(x \in X\), and \(B\) be a closed set such that \(x \notin B\).
• \(X - B\) is a neighbourhood of \(x\).
• Pick a neighbourhood \(U\) of \(x\) such that \(\overline{U} \subseteq X-B\).
• Let \(V = X - \overline{U}\)
• \(U\) and \(V\) are disjoint open sets.
• \(B \subseteq V\)
• If \(X\) is normal then, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\).
• Assume \(X\) is normal.
• Let \(A\) be a closed set and \(U\) an open set including \(A\)
• \(A\) and \(X - U\) are disjoint closed sets.
• Pick disjoint open sets \(V\) and \(W\) such that \(A \subseteq V\) and \(X - U \subseteq W\).
• \(\overline{V} \subseteq X - W \subseteq U\)
• If, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\), then \(X\) is normal.
• Assume that, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\).
• Let \(A\) and \(B\) be disjoint closed sets.
• Pick an open set \(U\) including \(A\) such that \(\overline{U} \subseteq X - B\).
• Let \(V\) = \(X - \overline{U}\)
• \(U\) and \(V\) are disjoint open sets.
• \(B \subseteq V\)
• \(\Box\)

Theorem 31.2

1. A subspace of a Hausdorff space is Hausdorff; a product of Hausdorff spaces is Hausdorff.
2. A subspace of a regular space is regular; a product of regular spaces is regular.

Proof

• A subspace of a Hausdorff space is Hausdorff.
• Let \(X\) be a Hausdorff space and \(Y \subseteq X\).
• Let \(x,y \in Y\) with \(x \neq y\).
• Pick disjoint open sets \(U\), \(V\) in \(X\) with \(x \in U\) and \(y \in V\).
• \(U \cap Y\) and \(V \cap Y\) are disjoint open sets in \(Y\) with \(x \in U \cap Y\) and \(y \in V \cap Y\).
• A product of Hausdorff spaces is Hausdorff.
• Let \((X_\alpha)\) be a family of Hausdorff spaces.
• Let \((x_\alpha),(y_\alpha) \in \prod_\alpha X_\alpha\) with \((x_\alpha) \neq (y_\alpha)\).
• Pick \(\alpha\) such that \(x_\alpha \neq y_\alpha\).
• Pick disjoint open sets \(U\), \(V\) in \(X_\alpha\) such that \(x_\alpha \in U\) and \(y_\alpha \in V\).
• \(\pi_\alpha^{-1}(U)\) and \(\pi_\alpha^{-1}(V)\) are disjoint open sets in \(\prod_\alpha X_\alpha\) with \((x_\alpha) \in \pi_\alpha^{-1}(U)\) and \((y_\alpha) \in \pi_\alpha^{-1}(V)\).
• A subspace of a regular space is regular.
• Let \(X\) be a regular space and \(Y \subseteq X\).
• \(Y\) is \(T_1\).
• For \(y \in Y\), we have \(\{y\} = \{y\} \cap Y\) is closed in \(Y\) because \(\{y\}\) is closed in \(X\).
• Let \(y \in Y\) and \(B\) be a closed set in \(Y\) such that \(y \notin B\).
• Pick a closed set \(C\) in \(X\) such that \(B = C \cap Y\).
• Pick disjoint open sets \(U\), \(V\) in \(X\) such that \(y \in U\) and \(C \subseteq V\).
• \(U \cap Y\) and \(V \cap Y\) are disjoint open sets in \(Y\) with \(y \in U \cap Y\) and \(B \subseteq V \cap Y\).
• A product of regular spaces is regular.
• Let \((X_\alpha)\) be a family of regular spaces.
• \(\prod_\alpha X_\alpha\) is \(T_1\).
• For \((x_\alpha) \in \prod_\alpha X_\alpha\) we have \(\overline{\{(x_\alpha)\}} = \overline{\prod_\alpha \{x_\alpha\}} = \prod_\alpha \overline{\{x_\alpha\}} = \prod_\alpha \{x_\alpha\} = \{(x_\alpha)\}\)
• Let \((x_\alpha) \in \prod_\alpha X_\alpha\) and \(U\) be a neighbourhood of \((x_\alpha)\).
• Pick \(\prod_\alpha U_\alpha\) such that each \(U_\alpha\) is open in \(X_\alpha\), \(U_\alpha = X_\alpha\) for all but finitely many \(\alpha\), \(x_\alpha \in U_\alpha\) and \(\prod_\alpha U_\alpha \subseteq U\).
• For each \(\alpha\) such that \(U_\alpha \neq X_\alpha\), pick \(V_\alpha\) open in \(X_\alpha\) such that \(x \in V_\alph\a\) and \(\overline{V_\alpha} \subseteq U_\alpha\). For the \(\alpha\) such that \(U_\alpha = X_\alpha\), let \(V_\alpha = X_\alpha\).
• \(\prod_\alpha V_\alpha\) is open in \(\prod_\alpha X_\alpha\) and we have \((x_\alpha) \in \prod_\alpha V_\alpha\) and \(\overline{\prod_\alpha V_\alpha} \subseteq \prod_\alpha U_\alpha\).
• \(\Box\)

Example 1

The space \(\mathbb{R}_K\) is Hausdorff but not regular.

To show it is Hausdorff, for any \(x,y \in \mathbb{R}_K\) with \(x < y\), we have \((-\infty, (x+y)/2\)\) and \(((x+y)/2, +\infty)\) are disjoint open sets that contain \(x\) and \(y\) respectively.

To show it is not regular, \(K\) is a closed set in \(\mathbb{R}_K\), but there are no disjoint open sets \(U\) and \(V\) with \(0 \in U\) and \(V \subseteq K\).

Example 2

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

• \(\mathbb{R}_l\) is \(T_1\).
• Let \(x \in \mathbb{R}\). Prove: \(\{x\}\) is closed.
• Let \(y \in \mathbb{R} - \{x\}\). Prove: there exists a neighbourhood of \(y\) that does not contain \(x\).
• If \(y < x\), then \([y,x)\) is a neighbourhood of \(y\) that does not contain \(x\).
• If \(y > x\), then \([y,y+1)\) is a neighbourhood of \(y\) that does not contain \(x\).
• For any disjoint closed sets \(A\) and \(B\), there exist disjoint open sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\).
• Let \(A\) and \(B\) be disjoint closed sets in \(\mathbb{R}_l\).
• For all \(a \in A\), choose \(x_a\) such that \([a,x_a) \cap B = \emptyset\).
• For all \(b \in B\), choose \(y_b\) such that \([b,y_b) \cap A = \emptyset\).
• Let \(U = \bigcup_{a \in A} [a,x_a)\).
• Let \(V = \bigcup_{b \in B} [b,y_b)\).
• \(U\) and \(V\) are open.
• \(U \cap V = \emptyset\)
• Assume for a contradiction \(t \in [a,x_a) \cap [b,y_b)\)
• Assume w.l.o.g. \(a < b\)
• \(b < x_a\)
• \(b \leq t < x_a\)
• This contradicts the fact that \([a,x_a) \cap B = \emptyset\)
• \(A \subseteq U\)
• \(B \subseteq V\)
• \(\Box\)

Example 3

The Sorgenfrey plane is not normal. Thus, not every regular space is normal, and the product of two normal spaces is not necessarily normal.

• Assume for a contradiction \(\mathbb{R}_l^2\) is normal.
• Let \(L = \{(x, -x): x \in \mathbb{R}\}\) as a subspace of \(\mathbb{R}_l^2\).
• \(L\) is closed in \(\mathbb{R}_l^2\).
• \(L\) is discrete.
• Every subset of \(L\) is closed in \(\mathbb{R}_l^2\).
• For every nonempty proper subset \(A\) of \(L\), choose disjoint open sets \(U_A\) and \(V_A\) including \(A\) and \(L-A\) respectively.
• Let \(D = \mathbb{Q}^2\).
• \(D\) is dense in \(\mathbb{R}_l^2\).
• Define \(\theta : \mathcal{P} L \rightarrow \mathcal{P} D\) by
\[ \begin{align} \theta(A) & = D \cap U_A & (\emptyset \neq A \neq L\) \\ \theta(\emptyset) & = \emptyset \\ \theta(A) & = D \end{align} \]
• \(\theta\) is injective.
• If \(A\) is a nonempty proper subset of \(L\) then \(\emptyset \neq D \cap U_A \neq D\)
• \(D \cap U_A \neq \emptyset\)
• Since \(U_A\) is a nonempty open set and \(D\) is dense.
• \(D \cap U_A \neq D\)
• Since \(D \cap V_A\) is nonempty and disjoint from \(U_A\).
• If \(A\) and \(B\) are distinct nonempty proper subsets of \(L\) then \(\theta(A) \neq \theta(B)\).
• Assume w.l.o.g. there exists \(x \in A\) such that \(x \notin B\).
• \(x \in U_A \cap V_B\)
• \(U_A \cap V_B\) is a nonempty open set.
• Pick a point \(d \in D \cap U_A \cap V_B\)
• \(d \in \theta(A)\) and \(d \notin \theta(B)\).
• This is a contradiction.
• Since \(|L| = 2^{\aleph_0}\) and \(|D| = \aleph_0\).
• \(\Box\)

Exercises

Exercise 1

Show that if \(X\) is regular, every pair of points of \(X\) have neighbourhoods whose closures are disjoint.

Solution

• Let \(X\) be a regular space.
• Let \(x,y \in X\) with \(x \neq y\)
• Pick disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(y \in V\).
• Pick an open set \(U'\) such that \(x \in U'\) and \(\overline{U'} \subseteq U\).
• Pick an open set \(V'\) such that \(y \in V'\) and \(\overline{V'} \subseteq V\).
• \(\overline{U'} \cap \overline{V'} = \emptyset\)
• \(\Box\)

Exercise 2

Show that if \(X\) is normal, every pair of disjoint closed sets have neighbourhoods whose closures are disjoint.

Solution

Similar.

Exercise 3

Show that every order topology is regular.

Solution

• Let \(X\) be a linearly ordered set under the order topology.
• Let \(x \in X\) and \(U\) be a neighbourhood of \(x\). Prove: there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\).
• Assume w.l.o.g. \(U\) is a basic open set.
• Case \(U = (a,b)\)
• Pick a point \(c\) such that \(a < c < x\) if such a point exists, or \(c = a\) if no such point exists.
• Pick a point \(d\) such that \(x < d < b\) if such a point exists, or \(c = b\) if no such point exists.
• Let \(V = (c,d)\)
• \(x \in V\) and \(\overline{V} \subseteq (a,b)\).
• Case \(U = [\bot, b)\) where \(\bot\) is least in \(X\).
• Pick a point \(d\) such that \(x < d < b\) if such a point exists, or \(c = b\) if no such point exists.
• Let \(V = [\bot,d)\)
• \(x \in V\) and \(\overline{V} \subseteq U\)
• Case \(U = (a,\top]\) where \(\top\) is greatest in \(X\).
• Pick a point \(c\) such that \(a < c < x\) if such a point exists, or \(c = a\) if no such point exists.
• Let \(V = (c,\top]\)
• \(x \in V\) and \(\overline{V} \subseteq U\)
• \(\Box\)

Exercise 4

Let \(\mathcal{T}\) and \(\mathcal{T}'\) be two topologies on the same set \(X\). Assume that \(\mathcal{T}' \supseteq \mathcal{T}\). If one of the topologies is Hausdorff (or regular, or normal), what does that imply about the other?

Solution

If \(\mathcal{T}\) is Hausdorff then \(\mathcal{T}'\) is Hausdorff. Let \(x,y \in X\) with \(x \neq y\). Pick disjoint sets \(U, V \in \mathcal{T}\) with \(x \in U\) and \(y \in V\). Then \(U\) and \(V\) are disjoint sets in \(\mathcal{T}'\).

The converse does not always hold. Take \(\mathcal{T}\) to be the indiscrete and \(\mathcal{T}'\) the discrete topology on a two-element set.

It is not true that, if \(\mathcal{T}\) is regular then \(\mathcal{T}'\) is regular. Take \(\mathcal{T}\) to be the standard topology and \(\mathcal{T}'\) the \(K\)-topology on \(\mathbb{R}\).

It is not true that, if \(\mathcal{T}'\) is regular then \(\mathcal{T}\) is regular. Take \(\mathcal{T}\) to be the indiscrete and \(\mathcal{T}'\) the discrete topology on a two-element set.

It is not true that, if \(\mathcal{T}\) is normal then \(\mathcal{T}'\) is normal. Take \(\mathcal{T}\) to be the standard topology and \(\mathcal{T}'\) the \(K\)-topology on \(\mathbb{R}\).

It is not true that, if \(\mathcal{T}'\) is normal then \(\mathcal{T}\) is normal. Take \(\mathcal{T}\) to be the indiscrete and \(\mathcal{T}'\) the discrete topology on a two-element set.

Exercise 5

Let \(f, g : X \rightarrow Y\) be continuous; assume that \(Y\) is Hausdorff. Show that \(E = \{x \in X \mid f(x) = g(x)\}\) is closed in \(X\).

Solution

• Let \(x \in X - E\). Prove: there exists a neighbourhood \(U\) of \(x\) such that \(U \subseteq X - E\).
• \(f(x) \neq g(x)\)
• Pick disjoint neighbourhoods \(V\), \(W\) of \(f(x)\) and \(g(x)\) respectively.
• Let \(U = f^{-1}(V) \cap g^{-1}(W)\)
• \(x \in U\)
• \(U\) is open
• \(U \subseteq X - E\)
• \(\Box\)

Exercise 6

Let \(p : X \rightarrow Y\) be a continuous surjective closed map. Show that if \(X\) is normal then so is \(Y\).

Solution

• For any \(y \in Y\) and any open set \(U\) in \(X\) that includes \(p^{-1}(y)\), there exists a neighbourhood \(W\) of \(y\) such that \(p^{-1}(W) \subseteq U\).
• Let \(y \in Y\) and let \(U\) be an open set in \(X\) that includes \(p^{-1}(y)\).
• \(p(X-U)\) is closed in \(Y\).
• Let \(W = Y - p(X-U)\)
• \(W\) is open
• \(y \in W\)
• \(p^{-1}(W) \subseteq U\)
• For any \(B \subseteq Y\) and any open set \(U\) in \(X\) that includes \(p^{-1}(B)\), there exists an open set \(W\) in \(Y\) that includes \(B\) such that \(p^{-1}(W) \subseteq U\).
• For all \(b \in B\), choose a neighbourhood \(W_b\) of \(b\) such that \(p^{-1}(W_b) \subseteq U\)
• Let \(W = \bigcup_{b \in B} W_b\)
• Let \(A\) and \(B\) be disjoint closed sets in \(Y\).
• \(p^{-1}(A)\) and \(p^{-1}(B)\) are disjoint closed sets in \(X\).
• Pick disjoint open sets \(U\) and \(V\) such that \(p^{-1}(A) \subseteq U\) and \(p^{-1}(B) \subseteq V\).
• Pick open sets \(U'\) and \(V'\) in \(Y\) such that \(A \subseteq U'\), \(B \subseteq V'\), \(p^{-1}(U') \subseteq U\) and \(p^{-1}(V') \subseteq V\).
• \(U'\) and \(V'\) are disjoint.
• \(\Box\)

Exercise 7

Let \(p : X \rightarrow Y\) be a closed continuous surjective map such that \(p^{-1}(y)\) is compact for each \(y \in Y\). (Such a map is called a perfect map.)

(a)

Show that if \(X\) is Hausdorff, then so is \(Y\).

Solution

• Assume \(X\) is Hausdorff.
• Let \(y,y' \in Y\) with \(y \neq y'\).
• \(p^{-1}(y)\) and \(p^{-1}(y')\) are disjoint compact sets in \(X\).
• Pick disjoint open sets \(U\) and \(V\) with \(p^{-1}(y) \subseteq U\) and \(p^{-1}(y') \subseteq V\).
• Pick neighbourhoods \(W_1\), \(W_2\) of \(y\) and \(y'\) such that \(p^{-1}(W_1) \subseteq U\) and \(p^{-1}(W_2) \subseteq V\).
• \(W_1 \cap W_2 = \emptyset\)

(b)

Show that if \(X\) is regular, so is \(Y\).

Solution

• Assume \(X\) is regular.
• Let \(y \in Y\) and \(B\) be a closed set in \(Y\) such that \(y \notin B\)
• \(p^{-1}(y)\) is compact and \(p^{-1}(B)\) is closed, and these two sets are disjoint.
• For all \(x \in p^{-1}(y)\), there exist disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(p^{-1}(B) \subseteq V\).
• \(\{U \text{ open in } X \mid \exists V \text{ open in } X. U \cap V = \emptyset, p^{-1}(B) \subseteq V\}\) is an open cover of \(p^{-1}(y)\).
• Pick a finite subcover \(\{U_1, \ldots, U_n\}\).
• For \(i = 1, \ldots, n\), pick \(V_i\) open with \(U_i \cap V_i = \emptyset\) and \(p^{-1}(B) \subseteq V_i\).
• Let \(U = U_1 \cup \cdots \cup U_n\) and \(V = V_1 \cap \cdots \cap V_n\).
• Pick neighbourhoods \(W\) and \(W'\) of \(y\) and \(B\) such that \(p^{-1}(W) \subseteq U\) and \(p^{-1}(W') \subseteq V\).
• \(W \cap W' = \emptyset\)
• \(\Box\)

(c)

Show that if \(X\) is locally compact, so is \(Y\).

Solution

• Assume \(X\) is locally compact.
• Let \(y \in Y\)
• For all \(x \in p^{-1}(y)\), there exists a neighbourhood \(U\) of \(x\) such that \(\overline{U}\) is compact.
• \(\{U \text{ open in } X \mid \overline{U} \text{ is compact}\}\) covers \(p^{-1}(y)\).
• Pick a finite subcover \(\{U_1, \ldots, U_n\}\).
• Let \(C = p(\overline{U_1}) \cup \cdots \cup p(\overline{U_n})\)
• \(C\) is compact.
• Pick a neighbourhood \(W\) of \(y\) such that \(p^{-1}(W) \subseteq U_1 \cup \cdots \cup U_n\)
• \(W \subseteq C\)
• \(\Box\)

(d)

Show that if \(X\) is second countable, then so is \(Y\).

Solution

• Assume \(X\) is second countable.
• Pick a countable basis \(\mathcal{B}\) for \(X\).
• For \(J \subseteq^\mathrm{fin} \mathcal{B}\), let \(V_J = \bigcup \{W \text{ open in } Y \mid p^{-1}(W) \subseteq \bigcup J\}\). Prove: \(\{V_J \mid J \subseteq^\mathrm{fin} \mathcal{B}\}\) is a basis for \(Y\).
• Let \(y \in Y\) and \(V\) be a neighbourhood of \(y\).
• For all \(x \in p^{-1}(y)\), there exists \(B \in \mathcal{B}\) such that \(x \in B \subseteq p^{-1}(V)\).
• Pick a finite \(J \subseteq \mathcal{B}\) such that \(\forall B \in J. B \subseteq p^{-1}(V)\) that covers \(p^{-1}(y)\).
• \(y \in V_J \subseteq V\)

Exercise 8

Let \(X\) be a topological space. Let \(G\) be a topological group. An action of \(G\) on \(X\) is a continuous map \(\alpha : G \times X \rightarrow X\) such that, denoting \(\alpha(g,x)\) by \(gx\), one has:

1. \(ex = x\) for all \(x \in X\)
2. \(g_1(g_2x) = (g_1g_2)x\) for all \(x \in X\) and \(g_1,g_2 \in G\)

Define \(x \sim gx\) for all \(x\) and \(g\). The resulting quotient space is denoted \(X/G\) and called the orbit space of the action \(\alpha\).

Theorem. Let \(G\) be a compact topological group. Let \(X\) be a topological space. Let \(\alpha\) be an action of \(G\) on \(X\). If \(X\) is Hausdorff, or regular, or normal, or locally compact, or second countable, so is \(X/G\).

Solution

By Exercises 6 and 7, it is enough to prove the canonical map \(p : X \rightarrow X / G\) is a perfect map. It is a quotient map by definition. For \(y \in X/G\) we have \(p^{-1}(y)\) is homeomorphic to \(G\) and so compact. It remains to show that \(p\) is a closed map.

• Let \(A \subseteq X\) be closed. Prove: \(p(A)\) is closed.
• \(GA\) is closed
• Let \(z \in X - GA\)
• \(\forall g \in G. gz \notin A\)
• For all \(g \in G\), pick neighbourhoods \(U_g\) of \(g\) and \(V_g\) and \(z\) such that \(U_g V_g \cap A = \emptyset\)
• Pick finitely many \(g_1, \ldots, g_n\) such that \(G = U_{g_1} \cup \cdots \cup U_{g_n}\).
• Let \(V = V_{g_1} \cap \cdots \cap V_{g_n}\)
• \(z \in V \subseteq X - GA\)
• \(p(A)\) is closed.
• \(\Box\)

Exercise 9

Let \(A = \{(x,-x) \mid x \in \mathbb{Q}\}\). Let \(B = \{(x,-x) \mid x \in \mathbb{R} - \mathbb{Q}\}\). If \(V\) is an open set of \(\mathbb{R}_l^2\) containing \(B\), show that there exists no open set \(U\) containing \(A\) that is disjoint from \(V\), as follows:

(a)

Let \(K_n\) be the set of all irrationals \(x \in [0,1]\) such that \([x,x+1/n) \times [-x,-x+1/n) \subseteq V\). Show that \([0,1]\) is the union of the sets \(K_n\) and countably many one-point sets.

Solution

• \([0,1] - \bigcup_n K_n \subseteq \mathbb{Q}\)
• Let \(x \in [0,1] - \mathbb{Q}\). Prove: \(x \in \bigcup_n K_n\).
• \((x,-x) \in V\)
• Pick basic open sets \([x,x+\epsilon)\) and \([-x,-x + \delta)\) such that \([x,x+\epsilon) \times [-x,-x+\delta) \subseteq V\).
• Pick \(n \in \mathbb{Z}_+\) such that \(1/n \leq \epsilon\) and \(1/n \leq \delta\).
• \(x \in K_n\)

(b)

Use Exercise 5 of §27 to show that some \(\overline{K_n}\) contains an open interval \((a,b)\) of \(\mathbb{R}\).

Solution

• Assume for a contradiction no \(\overline{K_n}\) contains an open interval.
• The sets \(\overline{K_n}\) and \(\{q\}\) for \(q \in [0,1] \cap \mathbb{Q}\) all have empty interior.
• \(\bigcup_n \overline{K_n} \cup \bigcup_q \{q\} = [0,1]\) has empty interior.
• Exercise 27.5
• This is a contradiction.
• \(\Box\)

(c)

Show that \(V\) includes the open parallelogram consisting of all points of the form \((x,-x+\epsilon)\) for which \(a < x < b\) and \(0 < \epsilon < 1/n\).

Solution

• Let \(a < x < b\) and \(0 < \epsilon < 1/n\).
• Let \(\delta = \min ((x-a)/2, \epsilon / 2)\)
• \(x - \delta \in \overline{K_n}\)
• Pick \(y \in (x-2\delta, x) \cap K_n\)
• We have \(y < x < y + 2\delta \leq y + \epsilon < y + 1/n\)
• \([y,y+1/n) \times [-y,-y+1/n) \subseteq V\)
• \(y \leq x < y+1/n\)
• \(-y \leq -x + \epsilon < -y+1/n\)
• \((x,-x+\epsilon) \in V\)
• \(\Box\)

(d)

Conclude that if \(q\) is a rational number with \(a < q < b\), then the point \((q,-q)\) of \(\mathbb{R}_l^2\) is a limit point of \(V\).

Solution

Given any basic open set \(B = [q,q+\delta_1) \times [-q,-q+\delta_2)\) of \((q,-q)\), pick \(\delta\) such that \(\delta < \delta_2\) and \(\delta < 1/n\). Then \((q,-q+\delta) \in B \cap V\).