Tietze Extension Theorem

Let $K$ be a normal Hausdorff space and $C \subseteq K$ be closed. Assume that the Urysohn lemma is a valid means to construct useful continuous maps.

Problem 1. For any $r > 0$ and any continuous map $f : C \to [-r, r]$ , construct a continuous map $g : K \to [-r/3, r/3]$ such that $(g-f)(C) \in [0, 2r/3]$ .

(Click for Solution)

Solution. Let $f : C \to [-r, r]$ be a continuous map. We observe that $K \backslash C$ is open. We will first construct a function $g$ that approximates $f$ reasonably well.

Decompose $[-r, r] = [-r,-r/3] \cup [-r/3,r/3] \cup [r/3, r]$ and define the closed subspaces

$C_1 := f^{-1}([-r,-r/3]), \quad C_2 := f^{-1}([-r/3,r/3]), \quad C_3 := f^{-1}([r/3,r])$

of $K$ . Since $C_1 \cap C_3 = \emptyset$ and $K$ is normal, the Urysohn lemma gives us a continuous function $g : K \to [-r/3,r/3]$ such that $g(C_1) = \{-r/3\}$ and $g(C_3) = \{r/3\}$ .

We observe the following estimates: by construction, $|g(x)| \leq r/3$ for any $x \in K$ . By considering cases, $|g(x) - f(x)| \leq 2r/3$ for any $x \in K$ .

Problem 2. For any continuous map $f : C \to [-1, 1]$ , construct one extension $F : K \to [-1, 1]$ of $f$ such that $F$ is continuous and $F|_C = f$ .

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Solution. Let $f : C \to [-1, 1]$ be any continuous map. Use Problem 1 to construct $g_1 : K \to [-1/3,1/3]$ such that

$|(f-g_1)(x)| \leq 2/3,\quad x \in C.$

Replacing $f$ with $f-g_1$ , use Problem 1 again to construct $g_2 : K \to [-1/3^2, 1/3^2]$ such that

$|(f - (g_1 + g_2))(x)| \leq (2/3)^2,\quad x \in C.$

Repeating the process, we obtain a sequence $\{g_n\}$ of continuous maps such that $g_n : K \to [-1/3^n, 1/3^n]$ and

$|(f - (g_1 + g_2 + \dots + g_n))(x)| \leq (2/3)^n,\quad x \in C.$

Then $f_n := g_1 + \cdots + g_n$ should converge to some limit function $F$ , and by our estimates, since $d_\infty( f_n, f) \leq (2/3)^n$ when restricted to $C$ , we have $f_n \to f$ uniformly on $C$ too. Therefore, $F|_C = f$ , as required.

Problem 3. For any continuous map $f : C \to \mathbb R$ , construct one extension $F : K \to \mathbb R$ of $f$ such that $F$ is continuous and $F|_C = f$ .

(Click for Solution)

Solution. Since $(-1, 1) \cong \mathbb R$ via the homeomorphism $t \mapsto t/(1-t^2)$ , it suffices to prove the result for $f : C \to (-1, 1)$ . By Problem 2, extend $f : C \to (-1, 1) \subseteq [-1, 1]$ to a continuous map $g : K \to [-1, 1]$ .

Define the closed set $D := g^{-1}(\{-1, 1\})$ . Since $g$ extends $f$ , $g(C) \subseteq (-1, 1)$ Hence, $C \cap D = \emptyset$ . Use Urysohn’s lemma to supply a continuous map $\phi : K \to [0, 1]$ where $\phi(D) = \{0\}$ and $\phi(C) = \{1\}$ .