Tietze extension by successive corrections

Try

A continuous function prescribed on a closed subset need not have a convenient formula outside it. Tietze’s theorem says that in a normal space, every continuous real-valued function on a closed subset can be extended continuously to the whole space.

For a bounded function, the proof does not guess the extension in one step. It builds continuous corrections whose sizes form a geometric series, making the remaining error on the closed subset shrink to zero.

Predict

Predict why one Urysohn separator is insufficient for an arbitrary function with many values. A separator only prescribes two levels on two closed sets. It can approximate a general bounded function coarsely, and the same process can then be applied to the residual.

Predict what fails if the original subset is not closed. The function 1/x on (0,1] is continuous there but cannot have a finite continuous extension across zero in R.

Experiment

Let the initial bound on the absolute value be M.

Correction step	Residual bound before correction	Correction size bound
0	M	M/3
1	2M/3	2M/9
2	4M/9	4M/27

The correction bounds sum to M. The residual bounds tend to zero, providing both convergence on the whole space and exact agreement on the original closed subset.

Understand

Suppose A is closed in normal X and |f|≤M on A. If M=0, the zero extension works. Otherwise the subsets of A where f≤−M/3 and f≥M/3 are disjoint closed subsets of X. Urysohn’s lemma gives a continuous h with values in [−M/3,M/3], taking the corresponding endpoints on those two sets. On either outer value region, subtracting that endpoint leaves error at most 2M/3. On the middle region, both f and h lie between −M/3 and M/3, so the same bound holds.

Apply this construction repeatedly to the residual on A. At step n its bound is M(2/3)ⁿ, while the correction has global bound one third of that. The correction series converges uniformly by the geometric majorant. Its sum F is continuous: for a given error tolerance, a uniformly small tail and continuity of a finite partial sum control nearby values. This proof works on any topological domain. The total absolute bound is at most M, and the residual on A tends uniformly to zero, so F=f there.

For an unbounded f, compose with a homeomorphism φ:R→(−1,1), extend the bounded composite to g:X→[−1,1], and let D be the closed set where g reaches either endpoint. It misses A. Choose a Urysohn function μ equal to one on A and zero on D. Then μg lies strictly between minus one and one everywhere and still agrees with φ∘f on A. Applying φ⁻¹ gives the desired real-valued extension.

Apply

Check the middle-region estimate rather than relying only on the two prescribed level sets. This is where a separator becomes an approximation to all of f. Then verify that each residual remains continuous on A, so the construction can be repeated.

Repair “Uniform convergence follows because every point’s residual tends to zero.” Pointwise residual convergence on A says nothing about convergence elsewhere. The summable global bounds on the correction functions are what establish a uniformly convergent series on X.

Continue

Tietze is an extension theorem for continuous scalar data, not a theorem that arbitrary maps into every target extend. Optional depth: extend a map into a finite cube coordinate by coordinate, and explain why the same argument does not automatically keep an extension inside an arbitrary nonlinear target subset.