Chapter 6 • Theory & Derivations

Unit 6: Normal Spaces (T4), Urysohn's Lemma & Tietze Extension Theorem

Higher separation axioms and functional analysis bridges: normal spaces ($T_4$), completely regular spaces ($T_{3.5}$, Tychonoff spaces), Pavel Urysohn's celebrated Lemma constructing continuous dyadic rational potentials, Heinrich Tietze's Extension Theorem extending continuous functions from closed subspaces, and the Urysohn Metrization Theorem embedding second-countable regular spaces into the Hilbert cube $[0, 1]^\mathbb{N}$.

§6.1 Normal Spaces & The $T_4$ Axiom

1. The Normal Space Axiom ($T_4$)

The highest classical separation axiom separates disjoint closed sets by disjoint open sets.

Definition 6.1 (Normal Space & $T_4$ Space):

A topological space $(X, \mathcal{T})$ is called normal if for every pair of disjoint closed subsets $A, B \subseteq X$ ($A \cap B = \emptyset$), there exist disjoint open sets $U, V \subseteq X$ such that:

A \subseteq U, \quad B \subseteq V, \qquad \text{and} \qquad U \cap V = \emptyset

A space is called a $T_4$ space if it is both normal and $T_1$.

Remark on Heredity: Unlike $T_0, T_1, T_2, T_3$, the normal property $T_4$ is NOT hereditary! A subspace of a normal space is not necessarily normal (e.g. the Sorgenfrey plane is not normal, but embeds as a subspace of a normal space). However, normality is closed-hereditary: every closed subspace of a normal space is normal!

2. Metric Spaces and Compact Hausdorff Spaces are Normal

Theorem 6.1:

Every metric space $(X, d)$ is normal ($T_4$).

Every compact Hausdorff space is normal ($T_4$).

3. Characterization via Open Neighborhoods of Closed Sets

Theorem 6.2 (Normal Neighborhood Characterization): A $T_1$ space $X$ is normal ($T_4$) if and only if for every closed set $A \subseteq X$ and every open set $W \supseteq A$, there exists an open set $U$ such that:

A \subseteq U \subseteq \operatorname{cl}(U) \subseteq W

Proof: ($\Rightarrow$) Let $A$ be closed and $A \subseteq W$ with $W$ open. Then $B = X \setminus W$ is closed, and $A \cap B = \emptyset$. By normality, there exist disjoint open sets $U, V$ with $A \subseteq U$ and $B \subseteq V$. Since $U \cap V = \emptyset$, $U \subseteq X \setminus V$. Since $X \setminus V$ is closed, $\operatorname{cl}(U) \subseteq X \setminus V$. Since $B \subseteq V$, $X \setminus V \subseteq X \setminus B = W$. Thus $A \subseteq U \subseteq \operatorname{cl}(U) \subseteq W$.

($\Leftarrow$) Let $A, B$ be disjoint closed sets. Then $W = X \setminus B$ is an open set containing $A$. By hypothesis, choose open $U$ with $A \subseteq U \subseteq \operatorname{cl}(U) \subseteq W$. Let $V = X \setminus \operatorname{cl}(U)$. Then $V$ is open, $B = X \setminus W \subseteq X \setminus \operatorname{cl}(U) = V$, and $U \cap V = \emptyset$. Thus $X$ is normal. $\blacksquare$

§6.2 Completely Regular Spaces ($T_{3.5}$, Tychonoff Spaces)

1. Separation by Continuous Functions

Andrey Tychonoff in 1930 introduced the intermediate separation axiom that bridges topology and functional analysis.

Definition 6.2 (Completely Regular / $T_{3.5}$ / Tychonoff Space):

A topological space $(X, \mathcal{T})$ is called completely regular if for every closed set $F \subseteq X$ and every point $x \notin F$, there exists a continuous function:

f: X \to [0, 1]

such that:

f(x) = 0 \quad \text{and} \quad f(y) = 1 \text{ for all } y \in F

A space is called a Tychonoff space (or $T_{3.5}$ space) if it is completely regular and $T_1$.

2. Position in the Separation Hierarchy

\text{Normal } (T_4) \implies \text{Tychonoff } (T_{3.5}) \implies \text{Regular } (T_3) \implies \text{Hausdorff } (T_2) \implies T_1 \implies T_0

Theorem 6.3 (Subspace and Product Preservation of $T_{3.5}$):

Complete regularity is hereditary: every subspace of a Tychonoff space is a Tychonoff space.

Complete regularity is product-invariant: arbitrary Cartesian products of Tychonoff spaces are Tychonoff spaces.

A topological space is Tychonoff if and only if it is homeomorphic to a subspace of a compact Hausdorff space (an embedding into a cube $[0, 1]^I$).

§6.3 Urysohn's Lemma: The Master Existence Theorem

1. Statement of Urysohn's Lemma

Pavel Urysohn proved in 1925 what is widely recognized as one of the most brilliant and fundamental theorems of general topology.

Theorem 6.4 (Urysohn's Lemma): Let $X$ be a normal space, and let $A$ and $B$ be two disjoint closed subsets of $X$ ($A \cap B = \emptyset$). Then there exists a continuous function:

f: X \to [0, 1]

such that:

f(x) = 0 \quad \text{for all } x \in A, \qquad \text{and} \qquad f(x) = 1 \quad \text{for all } x \in B

2. Full Line-by-Line Proof of Urysohn's Lemma

Proof: The proof proceeds by constructing a nested family of open sets indexed by the dyadic rational numbers in $[0, 1]$.

Step 1: Dyadic Rationals Indexing. Let $\mathbb{D} = \{m / 2^k : k \in \mathbb{N}_0, \; 0 \le m \le 2^k\}$ be the set of dyadic rationals in $[0, 1]$. $\mathbb{D}$ is countably infinite and dense in $[0, 1]$. Arrange the dyadic rationals in a sequence $r_0, r_1, r_2, \dots$ starting with $r_0 = 1$ and $r_1 = 0$.

Step 2: Constructing the Open Set Chain $\{U_r\}_{r \in \mathbb{D}}$. We construct for each $r \in \mathbb{D}$ an open set $U_r \subseteq X$ such that:

p < q \implies \operatorname{cl}(U_p) \subseteq U_q

For $r = 1$: let $U_1 = X \setminus B$. Since $B$ is closed, $U_1$ is open, and $A \subseteq U_1$.

For $r = 0$: $A$ is closed and $A \subseteq U_1$. By Theorem 6.2 (normality), choose open $U_0$ such that:

A \subseteq U_0 \subseteq \operatorname{cl}(U_0) \subseteq U_1

Inductive Construction: Let $\mathbb{D}_n = \{m / 2^n : 0 \le m \le 2^n\}$.

Assume $U_r$ has been defined for all $r \in \mathbb{D}_{n-1}$ satisfying the condition $\operatorname{cl}(U_p) \subseteq U_q$ whenever $p < q$. For each new dyadic fraction $r = \frac{2m + 1}{2^n} \in \mathbb{D}_n \setminus \mathbb{D}_{n-1}$, its immediate neighbors in $\mathbb{D}_{n-1}$ are:

p = \frac{m}{2^{n-1}} = \frac{2m}{2^n} \quad \text{and} \quad q = \frac{m+1}{2^{n-1}} = \frac{2m+2}{2^n}

By induction, $\operatorname{cl}(U_p) \subseteq U_q$. By Theorem 6.2, choose open $U_r$ such that:

\operatorname{cl}(U_p) \subseteq U_r \subseteq \operatorname{cl}(U_r) \subseteq U_q

Repeating this for all $r \in \mathbb{D}_n$ completes the inductive step!

We have constructed open sets $U_r$ for every $r \in \mathbb{D}$ such that:

p < q \implies \operatorname{cl}(U_p) \subseteq U_q

For convenience, define $U_r = \emptyset$ for $r < 0$, and $U_r = X$ for $r > 1$.

Step 3: Defining the Function $f: X \to [0, 1]$. For every $x \in X$, define:

f(x) = \inf \{r \in \mathbb{D} : x \in U_r\}

If $x \in A$, then $x \in U_r$ for all $r \ge 0$, so $f(x) = 0$.

If $x \in B$, then $x \notin U_1 = X \setminus B$. Hence $x \notin U_r$ for any $r \le 1$, so $f(x) = 1$.

For all $x \in X$, $0 \le f(x) \le 1$.

Step 4: Proving Continuity of $f$. We prove that for any $a \in \mathbb{R}$, the sets $\{x : f(x) < a\}$ and $\{x : f(x) > a\}$ are open in $X$.

By definition of infimum:

f(x) < a \iff \exists r \in \mathbb{D} \text{ such that } r < a \text{ and } x \in U_r

Therefore:

\{x \in X : f(x) < a\} = \bigcup_{\substack{r \in \mathbb{D} \\ r < a}} U_r

Since each $U_r$ is open, this is an arbitrary union of open sets, which is open!

Next, we show:

f(x) > a \iff \exists s \in \mathbb{D} \text{ such that } s > a \text{ and } x \notin \operatorname{cl}(U_s)

If $f(x) > a$, choose dyadics $s, t$ such that $a < s < t < f(x)$.

Since $t < f(x)$, $x \notin U_t$. Since $\operatorname{cl}(U_s) \subseteq U_t$, we have $x \notin \operatorname{cl}(U_s)$.

Conversely, if $x \notin \operatorname{cl}(U_s)$, then $x \notin U_s$, so $f(x) \ge s > a$.

Therefore:

\{x \in X : f(x) > a\} = \bigcup_{\substack{s \in \mathbb{D} \\ s > a}} (X \setminus \operatorname{cl}(U_s))

Since each $\operatorname{cl}(U_s)$ is closed, $X \setminus \operatorname{cl}(U_s)$ is open. Hence $\{x : f(x) > a\}$ is an arbitrary union of open sets, which is open!

Since the open rays $(-\infty, a)$ and $(a, \infty)$ form a subbasis for the standard topology on $\mathbb{R}$, $f: X \to [0, 1]$ is continuous. $\blacksquare$

§6.4 The Tietze Extension Theorem

1. Statement of the Tietze Extension Theorem

Heinrich Tietze in 1915 proved that continuous functions on closed subspaces can always be extended continuously to the entire space!

Theorem 6.5 (Tietze Extension Theorem): Let $X$ be a normal space and let $A \subseteq X$ be a closed subset.

Bounded Formulation: Any continuous function $f: A \to [a, b]$ has a continuous extension:

F: X \to [a, b] \qquad \text{such that} \quad F|_A = f

Unbounded Formulation: Any continuous function $f: A \to \mathbb{R}$ has a continuous extension:

F: X \to \mathbb{R} \qquad \text{such that} \quad F|_A = f

Proof Outline (Urysohn Series Approximation): Without loss of generality, assume $[a, b] = [-1, 1]$. Define the disjoint closed subsets of $A$ (which are closed in $X$):

A_0 = \{x \in A : f(x) \le -1/3\}, \qquad B_0 = \{x \in A : f(x) \ge 1/3\}

By Urysohn's Lemma, there exists continuous $g_0: X \to [-1/3, 1/3]$ with $g_0(A_0) = -1/3$ and $g_0(B_0) = 1/3$. Then on $A$, the error is reduced:

|f(x) - g_0(x)| \le \frac{2}{3}, \qquad \forall x \in A

Inductively constructing functions $g_n: X \to \mathbb{R}$ such that:

|g_n(x)| \le \frac{1}{3} \left(\frac{2}{3}\right)^n \quad \text{and} \quad \left|f(x) - \sum_{k=0}^n g_k(x)\right| \le \left(\frac{2}{3}\right)^{n+1}

By the Weierstrass M-test, the series $F(x) = \sum_{n=0}^\infty g_n(x)$ converges uniformly on $X$. Hence $F: X \to [-1, 1]$ is continuous, and $F(x) = f(x)$ for all $x \in A$. $\blacksquare$

§6.5 The Urysohn Metrization Theorem

1. When is a Topological Space Metrizable?

A fundamental quest of topology is finding purely topological conditions that guarantee a space comes from a metric.

Theorem 6.6 (Urysohn Metrization Theorem, 1925): Every second-countable regular ($T_3$) space is metrizable. In fact, every such space can be topologically embedded into the Hilbert cube $I^\infty = [0, 1]^\mathbb{N}$.

Proof Strategy:

By Urysohn's Lemma, since $X$ is regular and second-countable (hence normal), for each pair of basic open sets $B_n, B_m$ with $\operatorname{cl}(B_n) \subseteq B_m$, construct a continuous function $f_{n,m}: X \to [0, 1]$ with $f_{n,m}(\operatorname{cl}(B_n)) = 0$ and $f_{n,m}(X \setminus B_m) = 1$.

The collection $\{f_{n,m}\}$ is countable; enumerate it as $\{f_k\}_{k=1}^\infty$.

Define the embedding map $e: X \to [0, 1]^\mathbb{N}$ by $e(x) = (f_k(x))_{k=1}^\infty$.

Equip $[0, 1]^\mathbb{N}$ with the metric $d(u, v) = \sum_{k=1}^\infty \frac{|u_k - v_k|}{2^k}$.

Prove that $e: X \to e(X)$ is a homeomorphism, thereby pulling back the metric $d$ onto $X$! $\blacksquare$

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 6.1: Normality of Metric Spaces via Distance Functions

Let $(X, d)$ be a metric space, and let $A, B \subseteq X$ be two non-empty disjoint closed subsets ($A \cap B = \emptyset$). For any point $x \in X$, let $d(x, A) = \inf_{a \in A} d(x, a)$.

Prove that the distance function $x \mapsto d(x, A)$ is Lipschitz continuous with Lipschitz constant $1$:

|d(x, A) - d(y, A)| \le d(x, y), \qquad \forall x, y \in X

Prove that $d(x, A) + d(x, B) > 0$ for all $x \in X$.
Define the function $f: X \to [0, 1]$ by:

f(x) = \frac{d(x, A)}{d(x, A) + d(x, B)}

Prove that $f$ is continuous, $f(x) = 0$ for all $x \in A$, and $f(x) = 1$ for all $x \in B$.

Use $f$ to construct explicit disjoint open sets $U$ and $V$ separating $A$ and $B$, proving $(X, d)$ is normal ($T_4$).

1. Lipschitz Continuity of $d(x, A)$

For any $x, y \in X$ and any $a \in A$:

d(x, a) \le d(x, y) + d(y, a)

Taking the infimum over all $a \in A$:

d(x, A) \le d(x, y) + d(y, A) \implies d(x, A) - d(y, A) \le d(x, y)

Reversing the roles of $x$ and $y$:

d(y, A) - d(x, A) \le d(y, x) = d(x, y) \implies -(d(x, A) - d(y, A)) \le d(x, y)

Combining both inequalities:

|d(x, A) - d(y, A)| \le d(x, y)

Thus $x \mapsto d(x, A)$ is Lipschitz continuous with Lipschitz constant $L = 1$.

2. Denominator is Strictly Positive

Since $A$ and $B$ are closed:

$d(x, A) = 0 \iff x \in \operatorname{cl}(A) = A$.
$d(x, B) = 0 \iff x \in \operatorname{cl}(B) = B$.

Since $A \cap B = \emptyset$, no point $x$ can belong to both $A$ and $B$. Therefore, it is impossible for both $d(x, A) = 0$ and $d(x, B) = 0$. At least one of the distances is strictly positive for every $x \in X$:

d(x, A) + d(x, B) > 0, \qquad \forall x \in X

3. Properties of $f(x)$

Continuity: Both the numerator $d(x, A)$ and denominator $d(x, A) + d(x, B)$ are continuous, and the denominator is strictly positive everywhere.

The quotient of continuous functions with non-zero denominator is continuous. Hence $f: X \to [0, 1]$ is continuous.

On Set $A$: For $x \in A$, $d(x, A) = 0$ and $d(x, B) > 0$.

Thus $f(x) = \frac{0}{0 + d(x, B)} = 0$.

On Set $B$: For $x \in B$, $d(x, B) = 0$ and $d(x, A) > 0$.

Thus $f(x) = \frac{d(x, A)}{d(x, A) + 0} = 1$.

4. Construction of Disjoint Open Neighborhoods

Define the sets:

U = f^{-1}\left(\left[0, \frac{1}{3}\right)\right) = \left\{ x \in X : f(x) < \frac{1}{3} \right\}

V = f^{-1}\left(\left(\frac{2}{3}, 1\right]\right) = \left\{ x \in X : f(x) > \frac{2}{3} \right\}

Since $f$ is continuous and $[0, 1/3)$ and $(2/3, 1]$ are open in the subspace $[0, 1]$, $U$ and $V$ are open in $X$.
For every $x \in A$, $f(x) = 0 < 1/3 \implies A \subseteq U$.
For every $x \in B$, $f(x) = 1 > 2/3 \implies B \subseteq V$.
If $z \in U \cap V$, then $f(z) < 1/3$ and $f(z) > 2/3$, which is impossible! Thus $U \cap V = \emptyset$.

Therefore, $(X, d)$ is normal ($T_4$). $\blacksquare$

Advanced Example 6.2: Construction of Smooth Urysohn Bump Functions with Compact Support

In Euclidean space $\mathbb{R}^n$, construct an explicit smooth ($C^\infty$) Urysohn bump function $f: \mathbb{R}^n \to [0, 1]$ such that:

$f(x) = 1$ for all $\|x\| \le 1$ (on the closed unit ball $\bar{B}_1$).
$f(x) = 0$ for all $\|x\| \ge 2$ (outside the closed ball $\bar{B}_2$).
$f$ is infinitely differentiable ($C^\infty$) on all of $\mathbb{R}^n$.

1. The Standard Smooth Transition Function

Define the classic Cauchy flat function $h: \mathbb{R} \to \mathbb{R}$:

h(t) = \begin{cases} e^{-1/t} & \text{if } t > 0 \\ 0 & \text{if } t \le 0 \end{cases}

By standard calculus, $h$ is infinitely differentiable ($C^\infty$) on all of $\mathbb{R}$, with all derivatives vanishing at $t = 0$: $h^{(k)}(0) = 0$ for all $k \ge 1$.

2. Constructing the 1D Transition Function

Now define $g: \mathbb{R} \to [0, 1]$ by:

g(t) = \frac{h(2 - t)}{h(2 - t) + h(t - 1)}

Notice that for all $t \in \mathbb{R}$:

If $t \le 1$: $t - 1 \le 0 \implies h(t - 1) = 0$. Meanwhile $2 - t \ge 1 > 0 \implies h(2 - t) > 0$.

Thus $g(t) = \frac{h(2 - t)}{h(2 - t) + 0} = 1$.

If $t \ge 2$: $2 - t \le 0 \implies h(2 - t) = 0$. Meanwhile $t - 1 \ge 1 > 0 \implies h(t - 1) > 0$.

Thus $g(t) = \frac{0}{0 + h(t - 1)} = 0$.

For $1 < t < 2$: both $2 - t > 0$ and $t - 1 > 0$, so $h(2 - t) > 0$ and $h(t - 1) > 0$.

The denominator is strictly positive everywhere, so $g(t) \in (0, 1)$ is smooth.

Hence $g: \mathbb{R} \to [0, 1]$ is a smooth function with $g(t) = 1$ for $t \le 1$ and $g(t) = 0$ for $t \ge 2$.

3. Radial Extension to $\mathbb{R}^n$

For $x \in \mathbb{R}^n$, define the radial function:

f(x) = g(\|x\|^2)

For $\|x\| \le 1$: $\|x\|^2 \le 1 \implies f(x) = g(\|x\|^2) = 1$.
For $\|x\| \ge \sqrt{2} \approx 1.414$ (or setting $g(t)$ to scale between $1$ and $4$ for $r \in [1, 2]$):

Setting $f(x) = g(\|x\|)$ or $g(\|x\|^2 / 2)$: Specifically, setting $f(x) = \frac{h(4 - \|x\|^2)}{h(4 - \|x\|^2) + h(\|x\|^2 - 1)}$:

If $\|x\| \le 1$, $\|x\|^2 \le 1 \implies f(x) = 1$.
If $\|x\| \ge 2$, $\|x\|^2 \ge 4 \implies f(x) = 0$.
Since $x \mapsto \|x\|^2 = \sum_{i=1}^n x_i^2$ is smooth on $\mathbb{R}^n$, the composition $f$ is $C^\infty(\mathbb{R}^n)$.

This provides an explicit smooth Urysohn bump function with compact support in $\bar{B}_2(0)$.

Honors / Proof Challenge Example 6.3: Complete Proof of the Dyadic Continuity in Urysohn's Lemma

Provide the complete, rigorous mathematical proof that the potential function:

f(x) = \inf \{r \in \mathbb{D} : x \in U_r\}

constructed in Urysohn's Lemma is continuous:

Prove that for any $a \in \mathbb{R}$, $f(x) < a \iff \exists r \in \mathbb{D} \text{ with } r < a \text{ such that } x \in U_r$.
Prove that for any $a \in \mathbb{R}$, $f(x) > a \iff \exists s \in \mathbb{D} \text{ with } s > a \text{ such that } x \notin \operatorname{cl}(U_s)$.
Deduce that the preimages $f^{-1}((-\infty, a))$ and $f^{-1}((a, \infty))$ are open in $X$, completing the proof that $f: X \to [0, 1]$ is continuous.

1. Characterization of the Sublevel Sets $\{x : f(x) < a\}$

Let $a \in \mathbb{R}$. We prove that:

f(x) < a \iff x \in \bigcup_{\substack{r \in \mathbb{D} \\ r < a}} U_r

($\Rightarrow$) Suppose $f(x) < a$. By definition, $f(x) = \inf \{r \in \mathbb{D} : x \in U_r\}$. By the definition of the infimum, there exists some $r_0 \in \mathbb{D}$ such that $x \in U_{r_0}$ and $r_0 < a$. Therefore, $x \in \bigcup_{r < a} U_r$.

($\Leftarrow$) Suppose $x \in \bigcup_{r < a} U_r$. Then there exists some $r_0 \in \mathbb{D}$ with $r_0 < a$ such that $x \in U_{r_0}$. By definition of $f(x)$ as the infimum of all such values:

f(x) = \inf \{r \in \mathbb{D} : x \in U_r\} \le r_0 < a

Thus $f(x) < a$.

Since each $U_r$ is open in $X$, the union:

f^{-1}((-\infty, a)) = \{x \in X : f(x) < a\} = \bigcup_{\substack{r \in \mathbb{D} \\ r < a}} U_r

is an arbitrary union of open sets, which is open in $X$.

2. Characterization of the Superlevel Sets $\{x : f(x) > a\}$

Let $a \in \mathbb{R}$. We prove that:

f(x) > a \iff x \in \bigcup_{\substack{s \in \mathbb{D} \\ s > a}} (X \setminus \operatorname{cl}(U_s))

($\Rightarrow$) Suppose $f(x) > a$. Since the dyadic rationals $\mathbb{D}$ are dense in $\mathbb{R}$, choose two dyadic rationals $s, t \in \mathbb{D}$ such that:

$$a < s < t < f(x)$$

Since $t < f(x) = \inf \{r \in \mathbb{D} : x \in U_r\}$, the point $x$ cannot belong to $U_t$ (otherwise $f(x) \le t$). Thus $x \notin U_t$. Recall the fundamental property of the dyadic chain:

s < t \implies \operatorname{cl}(U_s) \subseteq U_t

Since $x \notin U_t$, we must have $x \notin \operatorname{cl}(U_s)$. Therefore:

x \in X \setminus \operatorname{cl}(U_s)

Since $s \in \mathbb{D}$ and $s > a$, $x \in \bigcup_{s > a} (X \setminus \operatorname{cl}(U_s))$.

($\Leftarrow$) Suppose $x \in \bigcup_{s > a} (X \setminus \operatorname{cl}(U_s))$. Then there exists $s \in \mathbb{D}$ with $s > a$ such that $x \notin \operatorname{cl}(U_s)$. Since $U_s \subseteq \operatorname{cl}(U_s)$, $x \notin U_s$. Furthermore, for any $r \in \mathbb{D}$ with $r \le s$, we have $U_r \subseteq U_s$, so $x \notin U_r$. Therefore, any $r \in \mathbb{D}$ for which $x \in U_r$ must satisfy $r > s$. Taking the infimum over all such $r$:

f(x) = \inf \{r \in \mathbb{D} : x \in U_r\} \ge s > a

Thus $f(x) > a$.

Since each $\operatorname{cl}(U_s)$ is a closed set in $X$, the complement $X \setminus \operatorname{cl}(U_s)$ is open in $X$. The union of open sets:

f^{-1}((a, \infty)) = \{x \in X : f(x) > a\} = \bigcup_{\substack{s \in \mathbb{D} \\ s > a}} (X \setminus \operatorname{cl}(U_s))

is therefore open in $X$.

3. Conclusion of Continuity

The collection of open rays:

\mathcal{S} = \{(-\infty, a) : a \in \mathbb{R}\} \cup \{(a, \infty) : a \in \mathbb{R}\}

forms a subbasis for the standard topology on $\mathbb{R}$. Since the preimage under $f$ of every subbasis element is open in $X$ (as established in Parts 1 and 2), by Theorem 3.1(7), the function:

f: X \to [0, 1]

is continuous. $\blacksquare$