Uniform continuity and compact domains

Try

The function f(x)=4x is uniformly continuous on R because |f(x)−f(y)|=4|x−y|. For any ε, δ=ε/4 works for every pair x,y. In contrast, a continuity proof for x² at a used a radius depending on |a|. That dependence does not by itself prove uniform continuity fails; it only shows the particular proof was point-dependent. Failure needs a counterexample defeating every shared radius.

Predict

Predict what happens to x² at large positive inputs. Compare x_n=n with y_n=n+1/n. Their distance is 1/n, tending to zero, while y_n²−x_n²=2+1/n². Thus no shared δ can force all output errors below 1. The local slopes become arbitrarily large, and the paired-input construction turns that geometric observation into a quantified disproof.

Experiment

Separate the conditions in a table.

Domain and function	One useful estimate	Uniform?
x² on [−2,2]	|x²−y²|≤4|x−y|	Yes
x² on R	Close pairs above stay separated in output	No
x on R	|x−y| exactly	Yes

Compactness is sufficient for continuous functions, not necessary. The last row shows that an unbounded domain does not automatically prevent uniform continuity. The relevant issue is whether one tolerance controls all pairs, not merely the domain’s appearance.

Understand

Uniform continuity on D means: for every ε>0 there is δ>0 such that all x,y∈D with |x−y|<δ satisfy |f(x)−f(y)|<ε. Ordinary continuity allows the radius to depend on the base point. A Lipschitz estimate |f(x)−f(y)|≤K|x−y| with finite K gives uniform continuity immediately; it is a sufficient condition, not an equivalent definition.

Here is the compact-domain proof. Assume continuous f on compact K is not uniformly continuous. Then some ε₀>0 admits pairs x_n,y_n∈K with |x_n−y_n|<1/n but output separation at least ε₀. Compactness extracts x_(n_k)→a∈K. The triangle inequality also gives y_(n_k)→a. Continuity makes both output subsequences tend to f(a), so their difference tends to zero, contradicting the fixed separation. Every hypothesis has a visible role: compactness extracts and retains the limit; continuity controls the outputs there.

Apply

Repair “continuous on a bounded interval implies uniformly continuous.” The interval must be compact or another argument must be supplied. On (0,1), f(x)=1/x is continuous but not uniformly continuous. For n≥2, choose x_n=1/n and y_n=1/(n+1). Their distance tends to zero while their outputs differ by 1. The missing endpoint permits increasingly steep behavior.

Also avoid the converse mistake: a uniformly continuous function need not be differentiable. The absolute-value function satisfies ||x|−|y||≤|x−y| on R but has no derivative at zero. Uniform continuity controls output changes without requiring a linear approximation.

Transfer: fill a missing endpoint without guessing

Uniform continuity has a useful consequence beyond choosing a shared radius. Let D be a nonempty subset of a metric space X, let Y be a complete metric space, and let f:D→Y be uniformly continuous. Then f has a unique continuous extension to the closure of D in X, and that extension is uniformly continuous.

Construct the value. For x in the closure choose xₙ∈D with d(xₙ,x)→0. The sequence is Cauchy. Uniform continuity sends it to a Cauchy sequence f(xₙ), which has a limit in Y by completeness. Define F(x) to be that limit. To show independence of the chosen sequence, take yₙ→x as well. Since d(xₙ,yₙ)→0, uniform continuity gives d(f(xₙ),f(yₙ))→0, so their limits agree. For x∈D the constant sequence shows F(x)=f(x).

Prove the shared tolerance survives. Given ε>0, choose δ from uniform continuity of f with output tolerance ε/2. If two closure points satisfy d(x,y)＜δ/3, choose approximating sequences from D. Their pairwise distances are eventually below δ. Taking limits of the output distances gives d(F(x),F(y))≤ε/2＜ε. Thus δ/3 works on the entire closure. Any other continuous extension has the same limits along these approximating sequences, proving uniqueness.

Worked endpoint. On (0,1), f(x)=√x satisfies |√x−√y|≤√|x−y|. For x≥y, square the left side and use (√x−√y)≤(√x+√y), giving (√x−√y)²≤x−y. The estimate proves uniform continuity and forces extension values F(0)=0 and F(1)=1. Sampling many points suggests these values; the inequality proves them.

Predict, then inspect the certificate

Changed case Missing endpoint with finite limitComplete codomain matters

f(x)=√x on (0,1). What value must a continuous extension assign at 0?

Your prediction Choose a prediction0Any valueNo extension exists Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Changed task. Which hypothesis fails for f(x)=1/x on (0,1)? Which fails for the identity Q→Q when extending to R? Give a concrete sequence in each argument, rather than just naming a theorem.

Feedback

For the reciprocal, inputs 1/n and 1/(n+1) approach each other while outputs differ by one: uniform continuity fails. For the rational identity, rational approximations to √2 are Cauchy but their images have no limit in Q: the codomain is not complete. Ordinary continuity on D alone does not guarantee either extension. These counterexamples diagnose different hypotheses.

Continue

Optional depth shows a uniformly continuous function sends Cauchy input sequences to Cauchy output sequences. Given ε, obtain a shared δ, then use the input Cauchy condition for that δ. Pointwise continuity cannot generally do this when the input limit lies outside the domain.

Save a positive proof with a shared tolerance and a negative proof with paired sequences. Explain why both inputs may vary in the counterexample. The next lesson uses compactness in a different way: instead of supplying a common radius, it ensures continuous functions attain extrema, while connected interval geometry forces intermediate values.