J. Munkres. Topology (2013) Chapter 1: Set Theory and Logic. 2: Functions

Definition 1: Function

A function is a triple \((A,B,f)\) where \(A\) and \(B\) are sets and \(f \subseteq A \times B\) is such that, for all \(x \in A\), there exists a unique \(y \in B\) such that \((x,y) \in f\). We often write just \(f\) for \((A,B,f)\). We call \(A\) the domain of \(f\) and \(B\) the codomain of \(f\). We write \(f : A \rightarrow B\) to mean \(f\) is a function with domain \(A\) and codomain \(B\).

The range or image of \(f\) is the set \(\{y \in B \mid \exists x \in A. (x,y) \in f \}\).

For \(x \in A\), we write \(f(x)\) for the unique \(y\) such that \((x,y) \in f\), and call this the value of \(f\) at \(x\).

Definition 2: Restriction

Let \(f : A \rightarrow B\) and \(C \subseteq A\). The restriction of \(f\) to \(C\) is the function \(f \restriction C : C \rightarrow B \) defined by \((f \restriction C)(x) = f(x)\) for \(x \in C\).

Definition 3: Composition

Given functions \(f : A \rightarrow B\) and \(g : B \rightarrow C\), the composite \(g \circ f : A \rightarrow C\) is the function given by \((g \circ f)(x) := g(f(x))\) for all \(x \in A\).

Definition 4: Injective

A function \(f : A \rightarrow B\) is injective or one-to-one iff, for all \(x,y \in A\), if \(f(x) = f(y)\), then \(x = y\).

Definition 5: Surjective

A function \(f: A \rightarrow B \) is surjective, or to map \(A\) onto \(B\), iff for all \(y \in B\) there exists \(x \in A\) such that \(f(x) = y\).

Definition 6: Bijective

A function \(f : A \rightarrow B\) is bijective or a one-to-one correspondence iff it is injective and surjective.

Proposition 7

The composite of two injective functions is injective.

Proof

Let \(f : A \rightarrow B\) and \(g : B \rightarrow C\) be injective. Let \(x,y \in A\). Assume \((g \circ f)(x) = (g \circ f)(y)\). Then \(g(f(x)) = g(f(y))\), so \(f(x) = f(y)\) because \(g\) is injective, and so \(x = y\) because \(f\) is injective. \(\Box\)

Proposition 8

The composite of two surjective functions is surjective.

Proof

Let \(f : A \rightarrow B\) and \(g : B \rightarrow C\) be surjective. Let \(z \in C\); we prove there exists \(x \in A\) such that \((g \circ f)(x) = z\). Pick \(y \in B\) such that \(g(y) = z\). Pick \(x \in A\) such that \(f(x) = y\). Then \((g \circ f)(x) = z\) as required. \(\Box\)

Proposition 9

The composite of two bijective functions is bijective.

Proof

From Propositions 8 and 9. \(\Box\)

Definition 10: Inverse

Let \(f : A \rightarrow B\) be bijective. Define the inverse \(f^{-1} : B \rightarrow A\) by: for \(y \in B\), we have \(f^{-1}(y)\) is the unique element in \(A\) such that \(f(f^{-1}(y)) = y\).

Proposition 11

Let \(f : A \rightarrow B\) be bijective. Then \(f^{-1} : B \rightarrow A\) is bijective.

Proof

We prove \(f^{-1}\) is injective. Let \(y,y' \in B\). Assume \(f^{-1}(y) = f^{-1}(y')\). Then \(y = f(f^{-1}(y)) = f(f^{-1}(y')) = y'\).

We now prove \(f^{-1}\) is surjective. Let \(x \in A\). Then \(f^{-1}(f(x)) = x\). \(\Box\)

Proposition 12

Let \(f : A \rightarrow B\). If there are functions \(g : B \rightarrow A\) and \(h : B \rightarrow A\) such that \(g(f(a)) = a\) for every \(a \in A\) and \(f(h(b)) = b\) for every \(b \in B\), then \(f\) is bijective and \(g = h = f^{-1}\).

Proof

We prove \(f\) is injective. Let \(x,y \in A\) and assume \(f(x) = f(y)\). Then \(x = g(f(x)) = g(f(y)) = y\).

We now prove \(f\) is surjective. For any \(y \in B\), there exists \(x \in A\) such that \(y = f(x)\), namely \(x = h(y)\).

For any \(y \in B\) we have \(h(y) = f^{-1}(y)\) because \(f(h(y)) = y\).

For any \(y \in B\) we also have \(g(y) = h(y)\) because \(g(y) = g(f(h(y))) = h(y)\). \(\Box\)

Definition 13: Image

Let \(f : A \rightarrow B\) and \(C \subseteq A\). The image of \(C\) under \(f\) is the set \(f(C) := \{f(x) \mid x \in C \}\).

Definition 14: Preimage

Let \(f : A \rightarrow B\) and \(C \subseteq B\). The preimage of \(C\) under \(f\) is the set \(f^{-1}(C) := \{x \in A \mid f(x) \in C \}\).

Proposition 15

Let \(f : A \rightarrow B\) be bijective and \(C \subseteq B\). Then the image of \(C\) under \(f^{-1}\) is equal to the preimage of \(C\) under \(f\).

Proof

\[ \begin{align} & x \text{ is in the image of } C \text{ under } f^{-1} \\ \Leftrightarrow & \exists y \in C. x = f^{-1}(y) \\ \Leftrightarrow & \exists y \in C. f(x) = y \\ \Leftrightarrow & f(x) \in C \\ \Leftrightarrow & x \text{ is in the preimage of } C \text{ under } f \end{align} \]

Proposition 16

Let \(f : A \rightarrow B\) and \(C, D \subseteq B\). Then \(f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)\).

Proof

\[ \begin{align} x \in f^{-1}(C \cap D) & \Leftrightarrow f(x) \in C \text{ and } f(x) \in D \\ & \Leftrightarrow x \in f^{-1}(C) \cap f^{-1}(D) & \Box \end{align} \]

Proposition 17

Let \(f : A \rightarrow B\) and \(C, D \subseteq B\). Then \(f^{-1}(C \cup D) = f^{-1}(C) \cup f^{-1}(D)\).

Proof

\[ \begin{align} x \in f^{-1}(C \cup D) & \Leftrightarrow f(x) \in C \text{ or } f(x) \in D \\ & \Leftrightarrow x \in f^{-1}(C) \cup f^{-1}(D) & \Box \end{align} \]

Proposition 18

Let \(f : A \rightarrow B\) and \(C \subseteq D \subseteq B\). Then \(f^{-1}(C) \subseteq f^{-1}(D)\).

Proof

If \(x \in f^{-1}(C)\) then \(f(x) \in C\) and so \(f(x) \in D\). Thus \(x \in f^{-1}(D)\). \(\Box\)

Proposition 19

Let \(f : A \rightarrow B\) and \(C, D \subseteq B\). Then \(f^{-1}(C - D) = f^{-1}(C) - f^{-1}(D)\).

Proof

\[ \begin{align} x \in f^{-1}(C - D) & \Leftrightarrow f(x) \in C \text{ or } f(x) \in D \\ & \Leftrightarrow x \in f^{-1}(C) \cup f^{-1}(D) & \Box \end{align} \]

Proposition 20

Let \(f : A \rightarrow B\) and \(C \subseteq D \subseteq A\). Then \(f(C) \subseteq f(D)\).

Proof

For all \(x \in C\) we have \(x \in D\) and so \(f(x) \in f(D)\). \(\Box\)

Proposition 21

Let \(f : A \rightarrow B\) and \(C, D \subseteq A\). Then \(f(C \cup D) = f(C) \cup f(D)\).

Proof

Given \(y \in B\), we have \(\exists x ((x \in C \vee x \in D) \wedge y = f(x))\) iff \(\exists x \in C. y = f(x)\) or \(\exists x \in D. y = f(x)\). \(\Box\)

Example 22

It is not true that, whenever \(f : A \rightarrow B\) and \(C, D \subseteq A\), then \(f(C \cap D) = f(C) \cap f(D)\). Let \(A = \{0,1\}\), \(B = \{0\}\), \(f = \{(0,0),(1,0)\}\), \(C = \{0\}\), \(D = \{1\}\). Then \(f(C \cap D) = f(\emptyset) = \emptyset\) but \(f(C) \cap f(D) = \{0\} \cap \{0\} = \{0\}\). \(\Box\)

Example 23

It is not true that, whenever \(f : A \rightarrow B\) and \(C, D \subseteq A\), then \(f(C - D) = f(C) - f(D)\). Let \(A = \{0,1\}\), \(B = \{0\}\), \(f = \{(0,0),(1,0)\}\), \(C = \{0,1\}\), \(D = \{1\}\). Then \(f(C - D) = f(\{0\}) = \{0\}\) but \(f(C) - f(D) = \{0\} - \{0\} = \emptyset\). \(\Box\)

Proposition 24

Let \(f : A \rightarrow B\) and \(C \subseteq A\). Then \(C \subseteq f^{-1}(f(C))\). Equality holds if \(f\) is injective.

Proof

For all \(x \in C\) we have \(f(x) \in f(C)\) and so \(x \in f^{-1}(f(C))\).

For the converse, assume \(f\) is injective and \(x \in f^{-1}(f(C))\). Then \(f(x) \in f(C)\), so there exists \(y \in C\) such that \(f(x) = f(y)\). Because \(f\) is injective we have \(x = y\) and so \(x \in C\) as required. \(\Box\)

Example 25

It is not true in general that, whenever \(f : A \rightarrow B\) and \(C \subseteq A\), then \(f^{-1}(f(C)) = C\). Define \(f : \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x) = 3x^2 + 2\). Let \(C = [0,1]\). Then \(f^{-1}(f(C)) = [-1,1] \neq C\).

Proposition 26

Let \(f : A \rightarrow B\) and \(C \subseteq B\). Then \(f(f^{-1}(C)) \subseteq C\). Equality holds if \(f\) is surjective.

Proof

Let \(y \in f(f^{-1}(C))\). Pick \(x \in f^{-1}(C)\) such that \(y = f(x)\). Then we have \(f(x) \in C\) and so \(y \in C\) as required.

For the converse, assume \(f\) is surjective. Let \(y \in C\). Pick \(x \in A\) such that \(y = f(x)\). Then \(x \in f^{-1}(C)\) and so \(y \in f(f^{-1}(C))\) as required.

Example 27

It is not true in general that, whenever \(f : A \rightarrow B\) and \(C \subseteq B\), then \(f(f^{-1}(C)) = C\). Define \(f : \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x) = 3x^2 + 2\). Let \(C = [0,5]\). Then \(f(f^{-1}(C)) = [2,5] \neq C\).

Exercise 1

Let \(f : A \rightarrow B\). Let \(A_0 \subseteq A\) and \(B_0 \subseteq B\).

(a)

Show that \(A_0 \subseteq f^{-1}(f(A_0))\) and that equality holds if \(f\) is injective.

Solution

See Proposition 24 above.

(b)

Show that \(f(f^{-1}(B_0)) \subseteq B_0\) and that equality holds if \(f\) is surjective.

Solution

See Proposition 26 above.

Exercise 2

Let \(f : A \rightarrow B\) and let \(A_i \subseteq A\) and \(B_i \subseteq B\) for \(i = 0\) and \(i = 1\). Show that \(f^{-1}\) preserves inclusions, unions, intersections, and differences of sets:

(a)

\[ B_0 \subseteq B_1 \Rightarrow f^{-1}(B_0) \subseteq f^{-1}(B_1) \]

Solution

See Proposition 18 above.

(b)

\[ f^{-1}(B_0 \cup B_1) = f^{-1}(B_0) \cup f^{-1}(B_1) \]

Solution

See Proposition 17 above.

(c)

\[ f^{-1}(B_0 \cap B_1) = f^{-1}(B_0) \cap f^{-1}(B_1) \]

Solution

See Proposition 16 above.

(d)

\[ f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1) \]

Solution

See Proposition 19 above.

Show that \(f\) preserves inclusions and unions only:

(e)

\[ A_0 \subseteq A_1 \Rightarrow f(A_0) \subseteq f(A_1) \]

Solution

See Proposition 20 above.

(f)

\[ f(A_0 \cup A_1) = f(A_0) \cup f(A_1) \]

Solution

See Proposition 21 above.

(g)

\( f(A_0 \cap A_1) \subseteq f(A_0) \cap f(A_1) \); show that equality holds if \(f\) is injective.

Solution

For all \(x \in A_0 \cap A_1\) we have \(f(x) \in f(A_0)\) (because \(x \in A_0\)) and \(f(x) \in f(A_1)\) (because \(x \in A_1\)), hence \(f(x) \in f(A_0) \cap f(A_1)\). Thus \(f(A_0 \cap A_1) \subseteq f(A_0) \cap f(A_1)\).

Assume \(f\) is injective. Let \(y \in f(A_0) \cap f(A_1)\). Pick \(x \in A_0\) such that \(y = f(x)\) and \(x' \in f(A_1)\) such that \(y = f(x')\). By injectivity, we have \(x = x'\) and so \(x \in A_0 \cap A_1\). Hence \(y \in f(A_0 \cap A_1)\) as required.

(h)

\(f(A_0 - A_1) \supseteq f(A_0) - f(A_1) \); show that equality holds if \(f\) is injective.

Solution

Let \(y \in f(A_0) - f(A_1)\). Pick \(x \in A_0\) such that \(y = f(x)\). Then \(x \notin A_1\) (lest \(y \in f(A_1)\)), so \(x \in A_0 - A_1\). Hence \(y \in f(A_0 - A_1)\) as required.

Now, assume \(f\) is injective. Let \(y \in f(A_0 - A_1)\). Pick \(x \in A_0 - A_1\) such that \(y = f(x)\). Then \(x \in A_0\) so \(y \in f(A_0)\); we must prove \(y \notin f(A_1)\). Assume for a contradiction \(y \in f(A_1)\). Pick \(x' \in A_1\) such that \(y = f(x')\). By injectivity, \(x = x'\), and so \(x \in A_1\), contradicting the fact that \(x \in A_0 - A_1\). Thus \(y \in f(A_0) - f(A_1)\) as required.

Exercise 3

Show that (b), (c), (f) and (g) of Exercise 2 hold for arbitrary unions and intersections.

Solution

(b)

Let \(f : A \rightarrow B\) and \(\mathcal{B}\) be a set of subsets of \(B\). We must show that \(f^{-1}\left( \bigcup \mathcal{B} \right) = \bigcup_{B_0 \in \mathcal{B}} f^{-1}(B_0)\).

\[ \begin{align} x \in f^{-1} \left( \bigcup \mathcal{B} \right) & \Leftrightarrow f(x) \in \bigcup \mathcal{B} \\ & \Leftrightarrow \exists B_0 \in \mathcal{B}. f(x) \in B_0 \\ & \Leftrightarrow \exists B_0 \in \mathcal{B}. x \in f^{-1}(B_0) \\ & \Leftrightarrow x \in \bigcup_{B_0 \in \mathcal{B}} f^{-1}(B_0) \end{align} \]

(c)

Let \(f : A \rightarrow B\) and \(\mathcal{B}\) be a nonempty set of subsets of \(B\). We must show that \(f^{-1} \left( \bigcap \mathcal{B} \right) = \bigcap_{B_0 \in \mathcal{B}} f^{-1}(B_0) \).

\[ \begin{align} x \in f^{-1} \left( \bigcap \mathcal{B} \right) & \Leftrightarrow f(x) \in \bigcap \mathcal{B} \\ & \Leftrightarrow \forall B_0 \in \mathcal{B}. f(x) \in B_0 \\ & \Leftrightarrow \forall B_0 \in \mathcal{B}. x \in f^{-1}(B_0) \\ & \Leftrightarrow x \in \bigcap_{B_0 \in \mathcal{B}} f^{-1}(B_0) \end{align} \]

(f)

Let \(f : A \rightarrow B\) and \(\mathcal{A}\) be a set of subsets of \(A\). We must show that \(f \left( \bigcup \mathcal{A} \right) = \bigcup_{A_0 \in \mathcal{A}} f(A_0) \).

\[ \begin{align} y \in f^ \left( \bigcup \mathcal{A} \right) & \Leftrightarrow \exists x \in \bigcup \mathcal{A}. y = f(x) \\ & \Leftrightarrow \exists x. \exists A_0 \in \mathcal{A} (x \in A_0 \wedge y = f(x)) \\ & \Leftrightarrow \exists A_0 \in \mathcal{A}. \exists x \in A_0. y = f(x) \\ & \Leftrightarrow \exists A_0 \in \mathcal{A}. y \in f(A_0) \\ & \Leftrightarrow y \in \bigcup_{A_0 \in \mathcal{A}} f(A_0) \end{align} \]

(g)

Let \(f : A \rightarrow B\) and \(\mathcal{A}\) be a nonempty set of subsets of \(A\). We must show that \(f \left( \bigcap \mathcal{A} \right) \subseteq \bigcap_{A_0 \in \mathcal{A}} f(A_0) \) and equality holds if \(f\) is injective.

Let \(y \in f \left( \bigcap \mathcal{A} \right)\). Pick \( x \in \bigcap \mathcal{A} \) such that \(y = f(x)\). Let \(A_0 \in \mathcal{A}\); we must show that \(y \in f(A_0)\). We have \(x \in A_0\) and \(y = f(x)\) so \(y \in f(A_0)\) as required.

Conversely, assume \(f\) is injective. Let \(y \in \bigcap_{A_0 \in \mathcal{A}} f(A_0) \). Pick \(A_1 \in \mathcal{A}\) (since \(\mathcal{A}\) is nonempty). We have \(y \in f(A_1)\). Pick \(x \in A_1\) such that \(y = f(x)\). We shall prove that \(x \in \bigcap \mathcal{A}\). It will follow that \(y \in f \left( \bigcap \mathcal{A} \right)\) as required.

Let \(A_0 \in \mathcal{A}\); we must show \(x \in A_0\). We have \(y \in f(A_0)\). Pick \(x' \in A_0\) such that \(y = f(x')\). By injectivity, we have \(x = x'\), and so \(x \in A_0\) as required.

Exercise 4

Let \(f : A \rightarrow B\) and \(g : B \rightarrow C\).

(a)

If \(C_0 \subseteq C\), show that \((g \circ f)^{-1}(C_0) = f^{-1}(g^{-1}(C_0))\).

Solution

\[ \begin{align} x \in (g \circ f)^{-1}(C_0) & \Leftrightarrow g(f(x)) \in C_0 \\ & \Leftrightarrow f(x) \in g^{-1}(C_0) \\ & \Leftrightarrow x \in f^{-1}(g^{-1}(C_0)) \end{align} \]

(b)

If \(f\) and \(g\) are injective, show that \(g \circ f\) is injective.

Solution

See Proposition 7 above.

(c)

If \(g \circ f\) is injective, what can you say about the injectivity of \(f\) and \(g\)?

Solution

We must have \(f\) is injective. Let \(x,y \in A\) and assume \(f(x) = f(y)\). Then \(g(f(x)) = g(f(y))\), and so \(x = y\) since \(g \circ f\) is injective.

It is not necessarily true that \(g\) is injective. Take \(A = \{0\}\), \(B = \{0,1\}\) and \(C = \{0\}\). Take \(f = \{(0,0)\}\) and \(g = \{(0,0),(1,0)\}\). Then \(g \circ f = \{(0,0)\}\) is injective but \(g\) is not.

(d)

If \(f\) and \(g\) are surjective, show that \(g \circ f\) is surjective.

Solution

See Proposition 8 above.

(e)

If \(g \circ f\) is surjective, what can you say about the surjectivity of \(f\) and \(g\)?

Solution

We must have that \(g\) is surjective. Let \(c \in C\). Pick \(a \in A\) such that \(c = g(f(a))\). Then there exists \(b \in B\) such that \(g(b) = c\), namely \(b = f(a)\).

It is not necessarily the case that \(f\) is surjective. Take \(A = \{0\}\), \(B = \{0,1\}\) and \(C = \{0\}\). Take \(f = \{(0,0)\}\) and \(g = \{(0,0),(1,0)\}\). Then \(g \circ f = \{(0,0)\}\) is surjective but \(f\) is not.

(f)

Summarize your answers to (b)-(e) in the form of a theorem.

Solution(?)

I'm not entirely sure what Munkres had in mind here. The best I can do:

Theorem

Let \(f : A \rightarrow B\) and \(g : B \rightarrow C\).

  1. Assume \(g\) is injective. Then \(g \circ f\) is injective if and only if \(f\) is injective.
  2. Assume \(f\) is surjective. Then \(g \circ f\) is surjective if and only if \(g\) is surjective.

Exercise 5

In general, let us denote the identity function for a set \(C\) by \(\mathrm{id}_C\). That is, define \(\mathrm{id}_C : C \rightarrow C\) to be the function given by the rule \(\mathrm{id}_C(x) = x\) for all \(x \in C\). Given \(f : A \rightarrow B\), we say that a function \(g : B \rightarrow A\) is a left inverse for \(f\) if \(g \circ f = \mathrm{id}_A\); and we say that \(h : B \rightarrow A\) is a right inverse for \(f\) if \(f \circ h = \mathrm{id}_B\).

(a)

Show that if \(f\) has a left inverse, then \(f\) is injective; and if \(f\) has a right inverse, then \(f\) is surjective.

Solution

Assume \(g\) is a left inverse for \(f\). Let \(x,y \in A\) and assume \(f(x) = f(y)\). Then we have \(x = g(f(x)) = g(f(y)) = y\).

Now assume \(h\) is a right inverse for \(f\). Let \(y \in B\). Then we have \(y = f(h(y))\). So there exists \(x \in A\) such that \(y = f(x)\), namely \(x = h(y)\).

(b)

Give an example of a function that has a left inverse but no right inverse.

Solution

Let \(A = \{0\}\) and \(B = \{0,1\}\). Let \(f = \{(0,0)\}\). Then \(g = \{(0,0)\}\) is a left inverse to \(f\), but \(f\) has no right inverse because it is not surjective.

(c)

Give an example of a function that has a right inverse but no left inverse.

Solution

Let \(A = \{0,1\}\) and \(B = \{0\}\). Let \(f = \{(0,0),(1,0)\}\). Then \(h = \{(0,0)\}\) is a right inverse to \(f\), but \(f\) has no left inverse because it is not injective.

(d)

Can a function have more than one left inverse? More than one right inverse?

Solution

Here is a function with more than one left inverse. Let \(A = \{0,1\}\) and \(B = \{0,1,2\}\). Define \(f : A \rightarrow B\) by \(f(0) = 0\) and \(f(1) = 1\). Then \(\{(0,0),(1,1),(2,0)\}\) and \(\{(0,0),(1,1),(2,1)\}\) are both left inverses to \(f\).

Here is a function with more than one right inverse. Let \(A = \{0,1\}\) and \(B = \{0\}\). Define \(f : A \rightarrow B\) by \(f(0) = 0\) and \(f(1) = 0\). Then \(\{(0,0)\}\) and \(\{(0,1)\}\) are both right inverses to \(f\).

(e)

Show that if \(f\) has both a left inverse \(g\) and a right inverse \(h\), then \(f\) is bijective and \(g = h = f^{-1}\).

Solution

We have \(f\) is injective and surjective by part (a), so \(f\) is bijective.

For any \(y \in B\), we have \(g(y) = g(f(h(y))) = h(y)\). Thus, \(g = h\).

For any \(y \in B\), we have \(f(h(y)) = y = f(f^{-1}(y))\), and so \(h(y) = f^{-1}(y)\) since \(f\) is injective. Thus \(h = f^{-1}\).

Exercise 6

Let \(f : \mathbb{R} \rightarrow \mathbb{R}\) be the funciton \(f(x) = x^3 - x\). By restricting the domain and codomain of \(f\) appropriately, obtain from \(f\) a bijective function \(g\). Draw the graphs of \(g\) and \(g^{-1}\). (There are several possible choices for \(g\).)

Solution

Well, the simplest solution is to restrict the domain and codomain to be \(\emptyset\)...

A less trivial solution: take \(g : \{x \in \mathbb{R} : x \geq 1\} \rightarrow \{y \in \mathbb{R} : y \geq 0\}\) to be the restriction of \(f\) to these two sets.

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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. \(X\) 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 \(\overline{V} \subseteq U\). \(X\) 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 th...
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