Sequences and the limits of sequential criteria

Try

In a metric space, if x lies in the closure of A, choose a point of A within distance 1/n of x. The chosen sequence converges to x. This familiar argument relies on the countable collection of balls with radii 1/n.

A general topological space may have no countable family that refines all neighborhoods. In that setting, closure remains well-defined, but a sequence may fail to test every neighborhood condition.

Predict

Predict whether replacing metric balls by a countable local base repairs the proof. It does, after making the base nested through finite intersections.

Predict whether sequence preservation always implies continuity. It does when the source is first-countable. Without that assumption, a topology can make all convergent sequences so restrictive that they fail to detect a noncontinuous map.

Experiment

Compare the valid scope.

Statement	General spaces	First-countable spaces
Sequence from A converging to x implies x∈closure A	Yes	Yes
Every closure point comes from a sequence in A	Not always	Yes
Continuous maps preserve sequence limits	Yes	Yes
Preserving all sequence limits implies continuity	Not always	Yes

The source space’s local base is used in the continuity converse. No first-countability assumption on the target is needed there.

Understand

Let U₁,U₂,… be an open local base at x and set V_n to the intersection of the first n members. This is still a local base and is nested. If x∈closure A, choose a_n∈A∩V_n. Every neighborhood of x contains some U_k, so all a_n for n≥k lie in that neighborhood. Thus a_n→x.

For the continuity converse, suppose f is not continuous at x. Some open V around f(x) has no input neighborhood mapped wholly into V. Choose x_n in V_n with f(x_n) outside V. Then x_n→x, while f(x_n) cannot converge to f(x), proving failure of sequence preservation.

Now consider an uncountable set X with the cocountable topology: nonempty opens have countable complement. If x_n converges to x, remove the countable set of all sequence values different from x. The remaining set is an open neighborhood of x. Eventual membership forces x_n=x for all sufficiently large n. Conversely eventual equality always gives convergence.

Consequently the identity from cocountable R to discrete R preserves every convergent sequence. Yet it is not continuous: a discrete singleton has a singleton preimage, which is not cocountable-open. This supplies an explicit failure of the unrestricted sequential criterion, rather than merely warning that a proof does not work.

Apply

Let A be R without zero in the cocountable topology. Every neighborhood of zero meets A, so zero belongs to its closure. No sequence in A can converge to zero, because convergence would require eventual equality to zero.

Repair “No sequence approaches zero, so zero is not a closure point.” The closure test quantifies over neighborhoods, not sequences. Here the neighborhood test succeeds while every sequence test fails. The example also shows this space is not first-countable.

Continue

Nets replace the countable index set by a directed set, allowing one test for every neighborhood. Optional depth: prove every sequence limit of A belongs to closure A without any countability assumption. The proof uses only eventual membership and the fact that each neighborhood then contains at least one term from A.