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). Assuming dependent choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain), 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 $F{\mathrm{\mid }}_A=f$, without supplying a formula. The following formula gives such an extension without dependent choice:

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

• 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.