MATH.23:5.3 - A failed limit argument reveals a stronger condition

Consider the claim that a pointwise limit of continuous real functions is continuous. On [0,1], let f_n(x)=x^n for positive natural n. Each function is continuous. At any fixed x<1, x^n tends to zero; at x=1 it equals one. The limit f is zero below one and equals one at one, so it is discontinuous.

Locate the failed proof step. Pointwise convergence lets the approximation index depend on x. Continuity near a point needs control over nearby x together. Choosing a good approximation at the one point alone does not control its neighborhood.

This suggests a sufficient condition: uniform convergence. For every positive error allowance, one index makes all later approximations close to f at every point of the domain.

Work the proposed repair. Fix a point x0 and an error allowance e>0. Choose N so that the approximation error between f_N and f is less than e/3 everywhere. Continuity of f_N gives a neighborhood of x0 in which the difference between f_N(x) and f_N(x0) is less than e/3. The triangle inequality bounds the difference between f(x) and f(x0) by these three errors, hence by e. This proves continuity of f at x0.

The repaired theorem supplies a condition for transferring continuity. Here proof analysis discovered which quantifier change made the desired inference possible. A subsequent limit construction can use the proved condition to retain continuity.

Uniformity is sufficient, not necessary. On the domain [0,1), the same x^n sequence has the continuous limit zero, but convergence is not uniform: for every n there is an x<1 with x^n=1/2. The next question can therefore seek a weaker condition suited to the receiving use, rather than treating uniform convergence as the definition of every acceptable limit.