Sequential criteria for limits and continuity

Try

A continuous function carries every convergent input sequence to the corresponding output limit. For f(x)=x² and x_n→2, the product limit law gives f(x_n)→4. This offers a second language for continuity: instead of controlling all inputs in one neighborhood at once, inspect all possible sequences approaching the point. The word all is essential. One successful sequence is not enough.

Predict

Predict how to disprove continuity if the neighborhood definition fails. Failure supplies one output tolerance ε₀ that cannot be enforced by any input radius. Use radius 1/n at step n, and choose an input violating that tolerance. The chosen inputs approach a while their outputs stay separated from f(a). The same ε₀ persists through the construction; allowing it to shrink would not produce the required contradiction.

Experiment

Compare two input sequences for the step function f(x)=0 when x≥0 and f(x)=1 when x<0.

Inputs	Input limit	Output values	Match f(0)?
x_n=1/n	0	0	Yes
y_n=−1/n	0	1	No

The first row might look convincing in a one-sided sample. The second supplies a decisive witness. Draw both approaches on a number line and label the fixed output gap. Explain why the function’s assigned value at zero determines which sequence contradicts continuity.

Understand

Prove the sequential criterion in the forward direction. Suppose f is continuous at a and x_n→a with every x_n in its domain. Given ε>0, continuity provides δ>0 controlling the output. Sequence convergence provides N so all n≥N satisfy |x_n−a|<δ. Therefore all corresponding output errors are below ε.

For the converse, argue contrapositively. If continuity fails, there exists ε₀>0 such that for every δ>0 there is a domain point x with |x−a|<δ and |f(x)−f(a)|≥ε₀. Choose such a point x_n for δ=1/n. Then x_n→a but f(x_n) cannot tend to f(a), contradicting preservation of every convergent input sequence. This proof converts the negated quantifiers into a concrete witness sequence.

Apply

The indicator of Q is discontinuous at every real point. Rational and irrational density guarantee inputs of both types arbitrarily close to any a. If a is rational, choose irrational inputs converging to it; their outputs are zero while f(a)=1. If a is irrational, choose rational inputs; their outputs are one while f(a)=0. The argument works at each point without treating a finite plot as representative.

Repair “if f(x_n) converges, then x_n converges.” A constant function sends every input sequence to one constant output, including divergent input sequences. Continuity is a forward implication with a specified input limit, not an invertibility theorem. An inverse-continuity claim requires a suitable inverse and additional reasoning.

Continue

Optional depth states the analogous criterion for a punctured limit: test every sequence of domain points different from a that tends to a. The exclusion prevents the assigned value at a from contaminating a claim intended to concern only nearby values. The accumulation-point assumption ensures such sequences exist.

Write a proof twice, once with neighborhoods and once with sequences, and compare the quantifier choices. For a counterexample, preserve a fixed positive output gap along the entire witness sequence. Next, uniform continuity will demand one neighborhood tolerance for every base point simultaneously, a stronger condition that compact domains can supply.