Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality

Assume $X$ is normal, $A \subseteq X$ is closed, $a \le b$ are reals, and $f : A \to [a,b]$ is continuous.

Assume instead that $X$ is such that every continuous $g : C \to [p,q]$ on a closed $C \subseteq X$, $p \le q$ reals, extends continuously to $X \to [p,q]$.

Under step 1.1: if $a=b$ the constant map $F : X \to \{a\} \subseteq [a,b]$, $F \equiv a$, is continuous and $F|_A = f$, since $f : A \to \{a\}$ forces $f \equiv a$. Assume from here that $a<b$.

Under step 1.2: let $C, E \subseteq X$ be disjoint closed sets; $C \cup E$ is closed, and $C, E$ are each open in the subspace $C \cup E$, being the complement there of the other, which is closed. Define $k : C \cup E \to \{0,1\} \subseteq [0,1]$ by $k \equiv 0$ on $C$ and $k \equiv 1$ on $E$; $k$ is constant, hence continuous, on each of $C$ and $E$, so $k$ is continuous on $C \cup E$ by [L8].

Under steps 1.1 and 2.1: put $c := (a+b)/2$ and $r := (b-a)/2 > 0$, and define $f_0 : A \to \mathbb{R}$ by $f_0(x) := f(x)-c$; $f_0$ is continuous, being $f$ minus a constant, and $f_0[A] \subseteq [-r,r]$, since $f[A] \subseteq [a,b] = [c-r,c+r]$.

Under step 1.2: by hypothesis applied to the closed set $C \cup E$ and $p:=0, q:=1$, fix a continuous $K : X \to [0,1]$ with $K|_{C\cup E} = k$.

Under step 1.1: for $n \in \mathbb{N}$ put $M_n := r(2/3)^n$. Call a pair $(f_n,g_n)$, with $f_n : A \to \mathbb{R}$ and $g_n : X \to \mathbb{R}$ continuous, admissible at level $n$ when $|f_n(x)| \le M_n$ for $x \in A$; $|g_n(x)| \le M_n/3$ for $x \in X$; $g_n(x) = -M_n/3$ for $x \in A$ with $f_n(x) \le -M_n/3$; and $g_n(x) = M_n/3$ for $x \in A$ with $f_n(x) \ge M_n/3$.

Under step 1.2: by [L7], put $O_1 := K^{-1}(\,[0,\tfrac12)\,)$, $O_2 := K^{-1}(\,(\tfrac12,1]\,)$, open by [L3]. $C \subseteq O_1$, since $K \equiv 0 \in [0,\tfrac12)$ on $C$; $E \subseteq O_2$, since $K \equiv 1 \in (\tfrac12,1]$ on $E$; and $O_1 \cap O_2 = \varnothing$, the two target sets being disjoint.

Under step 1.1: put $A_0^- := \{x \in A : f_0(x) \le -M_0/3\}$, $A_0^+ := \{x \in A : f_0(x) \ge M_0/3\}$; both closed in $A$ by [L3] and hence in $X$ by [L2], and disjoint since $-M_0/3 < M_0/3$. By [L1] fix continuous $h_0 : X \to [0,1]$ with $A_0^- \subseteq h_0^{-1}(\{0\})$ and $A_0^+ \subseteq h_0^{-1}(\{1\})$, and put $g_0 := (M_0/3)(2h_0-1)$, continuous.

Under step 1.1: let $n \in \mathbb{N}$ and let $(f_n,g_n)$ be admissible at level $n$; define $f_{n+1} : A \to \mathbb{R}$ by $f_{n+1}(x) := f_n(x)-g_n(x)$, continuous.

Under step 1.2: since $C, E$ were an arbitrary disjoint closed pair, step 4.2 exhibits disjoint open supersets for every such pair, so $X$ is normal by [A2]; this is clause 2, and it uses [A1] nowhere.

Under step 1.1: $(f_0,g_0)$ is admissible at level $0$: $|f_0| \le M_0$ on $A$ by step 3.1; $|g_0(x)| = (M_0/3)|2h_0(x)-1| \le M_0/3$ for every $x$, since $h_0(x) \in [0,1]$; $g_0(x) = -M_0/3$ for $x \in A_0^-$, where $h_0(x)=0$; and $g_0(x)=M_0/3$ for $x \in A_0^+$, where $h_0(x)=1$.

Under step 1.1, continuing under step 5.2: for $x \in A$ with $f_n(x) \le -M_n/3$: $g_n(x)=-M_n/3$ (admissibility), so $f_{n+1}(x) = f_n(x)+M_n/3 \in [-2M_n/3,\,0]$, using $-M_n \le f_n(x) \le -M_n/3$; for $x \in A$ with $f_n(x) \ge M_n/3$: $f_{n+1}(x) = f_n(x)-M_n/3 \in [0,\,2M_n/3]$; for $x \in A$ with $-M_n/3 < f_n(x) < M_n/3$: $|g_n(x)| \le M_n/3$ gives $f_{n+1}(x) \in (-2M_n/3,\,2M_n/3)$. In every case $|f_{n+1}(x)| \le 2M_n/3 = M_{n+1}$.

Under step 1.1: put $A_{n+1}^- := \{x\in A: f_{n+1}(x)\le -M_{n+1}/3\}$, $A_{n+1}^+ := \{x\in A: f_{n+1}(x)\ge M_{n+1}/3\}$; closed in $X$ by [L2], [L3], and disjoint. By [L1] fix continuous $h_{n+1}:X\to[0,1]$ with $A_{n+1}^- \subseteq h_{n+1}^{-1}(\{0\})$, $A_{n+1}^+ \subseteq h_{n+1}^{-1}(\{1\})$, and put $g_{n+1} := (M_{n+1}/3)(2h_{n+1}-1)$.

Under step 1.1: $(f_{n+1},g_{n+1})$ is admissible at level $n+1$, by step 6.2 and the same computation as step 6.1 with $h_{n+1}, g_{n+1}, M_{n+1}$ in place of $h_0,g_0,M_0$. So every admissible pair at level $n$ has an admissible successor at level $n+1$.

Under step 1.1: put $P := \{\, (n,f_n,g_n) : n \in \mathbb{N},\ (f_n,g_n) \text{ admissible at level } n \,\}$, and for $(n,f,g),(n',f',g') \in P$ say $(n,f,g) \mathbin{R} (n',f',g')$ when $n'=n+1$ and $f' = (f-g)|_A$ pointwise. $P$ is nonempty by step 6.1, and $R$ is entire on $P$ by steps 5.2, 6.2, 6.3 and 7.1 (the pair produced there has $f_{n+1} = (f_n-g_n)|_A$ exactly as step 5.2 defines it). By [A1] with $a := (0,f_0,g_0)$, fix a sequence $\big((n_k,F_k,G_k)\big)_{k \in \mathbb{N}}$ with $(n_0,F_0,G_0)=(0,f_0,g_0)$ and $(n_k,F_k,G_k) \mathbin{R} (n_{k+1},F_{k+1},G_{k+1})$ for every $k$; as $R$ forces $n'=n+1$, induction gives $n_k=k$, so $(F_k,G_k)$ is admissible at level $k$ for every $k$, with $F_{k+1} = (F_k-G_k)|_A$.

Under step 1.1: by [L4], $\sum_n M_n/3 = r$, convergent; by [L5] applied to $(G_n)$ and $(M_n/3)$ (each $|G_n(x)| \le M_n/3$ for all $x$, by admissibility), for every $x \in X$ the series $\sum_n G_n(x)$ converges, and $F := \sum_{n=0}^{\infty} G_n$ is a continuous map $X \to \mathbb{R}$.

Under step 1.1: for $x \in A$ and $N \in \mathbb{N}$: by the telescoping of step 8.1, $\sum_{n<N} G_n(x) = F_0(x) - F_N(x) = f_0(x)-F_N(x)$, since $F_0=f_0$.

Under step 1.1: for every $x \in X$ and $N \in \mathbb{N}$, $\big|\sum_{n<N} G_n(x)\big| \le \sum_{n<N}|G_n(x)| \le \sum_{n<N} M_n/3 \le r$, by [L6] and admissibility; letting $N \to \infty$, since $\sum_{n<N}G_n(x) \to F(x)$ (step 9.1) and order is preserved in the limit ([L6]), $|F(x)| \le r$.

Under step 1.1: for $x \in A$: $|F_N(x)| \le M_N = r(2/3)^N \to 0$ as $N \to \infty$, by admissibility of $F_N$ (step 8.1) and [L4]; so by step 9.2, $\sum_{n<N} G_n(x) = f_0(x)-F_N(x) \to f_0(x)-0 = f_0(x)$.

Under step 1.1: for $x \in A$: $\sum_{n<N} G_n(x) \to F(x)$ by step 9.1 and $\to f_0(x)$ by step 10.2; since a real sequence has at most one limit ([L6]), $F(x) = f_0(x)$.

Under step 1.1: define $\hat F : X \to \mathbb{R}$ by $\hat F(x) := F(x)+c$, continuous; for $x \in X$, $\hat F(x) \in [c-r,c+r] = [a,b]$ by step 10.1; for $x \in A$, $\hat F(x) = F(x)+c = f_0(x)+c = f(x)$ by step 11.1 and the definition of $f_0$ in step 3.1.

Steps 2.1 and 12.1 show that, under the hypothesis of step 1.1, a continuous $F : X \to [a,b]$ with $F|_A=f$ exists — either the constant map of step 2.1 when $a=b$, or $\hat F$ of step 12.1 when $a<b$ — which is clause 1.

Steps 13.1 and 5.3 establish clauses 1 and 2 respectively.