A continuous function on $[0,1] \subseteq \mathbb{R}$ extended to all of $\mathbb{R}$, both by Tietze and by hand

Let $A := [0,1] \subseteq \mathbb{R}$, closed, and $f : A \to \mathbb{R}$, $f(x) := x^2$, continuous. $\mathbb{R}$ is normal (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal), so Under dependent choice, a continuous real-valued map on a closed subspace of a normal space extends to the whole space, and a map into an open interval extends into that same open interval guarantees a continuous $F : \mathbb{R} \to \mathbb{R}$ with $F|_A = f$, with no formula supplied. One is written down here directly:

$F(x) \;:=\; \big(\max\{0,\ \min\{1,\ x\}\}\big)^2.$

• The trick generalises. For any closed bounded interval $[a,b]$ and any continuous $f : [a,b] \to \mathbb{R}$, precomposing $f$ with the clamp $x \mapsto \max\{a,\min\{b,x\}\} : \mathbb{R} \to [a,b]$ gives a continuous extension of $f$ to all of $\mathbb{R}$, by the same two-step argument as above. Nothing about $x^2$ is used beyond its own continuity on $[0,1]$.

• This is not the extension the recursive proof of 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 would produce. That proof builds an extension as a uniformly convergent series of Urysohn functions, and it makes no claim of matching the clamp-composition extension above; both are legitimate continuous extensions, and neither is canonical.