False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: a sequentially continuous map between topological spaces is continuous

Statement

False claim: if $X$ and $Y$ are topological spaces and $f : X \to Y$ is sequentially continuous (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), then $f$ is continuous (Continuity of a map of topological spaces at a point and globally).

One half of the relation between the two notions is a theorem: continuity always implies sequential continuity, and, assuming the Axiom of Countable Choice, in a first countable source the converse holds as well (Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there). The claim above drops the first-countability hypothesis, and the witness is the identity map from $R$ with the cocountable topology to $R$ with its usual topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and it is exhibited in full below rather than cited, so that this page does not depend on its companion.

Facts & Assumptions

Given: The set $R$ carrying the cocountable topology $T_{c o c}$ on the one hand and its usual topology $T_{R}$ on the other, and the identity function $i d : (R, T_{c o c}) \to (R, T_{R})$ .

[A1]

In the cocountable topology on $R$ the open sets are $\emptyset$ together with the sets whose complement is at most countable (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Finite, countably infinite, countable, uncountable).

[A2]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b)); it is sequentially continuous at $p$ when every sequence converging to $p$ has its image converging to $f (p)$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L1]

$R$ with $d_{R} (x, y) = ∣ x - y ∣$ is a metric space whose metric topology is the usual topology of $R$ , and $B (x, r) = (x - r, x + r)$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open ball, closed ball and sphere in a metric space).

[L2]

Every ball is open in the metric topology (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[L3]

For $a < b$ in $R$ the open interval $(a, b)$ is uncountable (Every nondegenerate interval of $R$ is uncountable); every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L4]

A nonempty set admitting a surjection from $N$ is at most countable (A nonempty set is at most countable iff it is a surjective image of $N$ ).

[L5]

$0 < 1$ in $R$ (The multiplicative identity is positive), and adding $1$ to both sides of $0 < 1$ gives $1 < 1 + 1$ (Order is preserved by adding a constant and by adding inequalities).

[L6]

Every point lies in each of its neighbourhoods (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

Refutation

technique · direct

1.1

The radius $1$ is positive by [L5], so $V : = B (0, 1) = (- 1, 1)$ is a ball, and it is open in the usual topology of $R$ .

L1L2L5

1.2

By [L5] one has $1 < 1 + 1$ , so the interval $(1, 1 + 1)$ is uncountable by [L3]; and $(1, 1 + 1) \subseteq R ∖ (- 1, 1)$ , since $x > 1$ excludes $x < 1$ .

L1L3L5

1.3

Let $(x_{k})$ be a sequence in $R$ converging to $p$ in the cocountable topology, and let $R : = {x_{k} : k \in N}$ be its range; the map $k \mapsto x_{k}$ is a surjection $N \to R$ and $R \neq \emptyset$ , so $R$ is at most countable.

givenL4

2.1

$R ∖ (- 1, 1)$ is not at most countable: otherwise its subset $(1, 1 + 1)$ would be at most countable by [L3], contradicting step 1.2. Hence $V = (- 1, 1)$ is nonempty and its complement is not at most countable, so $V \notin T_{c o c}$ .

step 1.2A1L3

2.2

With $R$ as in step 1.3, the set $S : = R ∖ {p}$ is at most countable by [L3], so $U : = R ∖ S$ is open in the cocountable topology by [A1], and $p \in U$ .

step 1.3A1L3

3.1

${i d}^{- 1} [V] = V$ , which is open in the usual topology by step 1.1 and not open in the cocountable topology by step 2.1; so $i d$ is not continuous.

step 1.1step 2.1A2

3.2

$U$ is a neighbourhood of $p$ in the cocountable topology by step 2.2, so convergence gives $K \in N$ with $x_{k} \in U$ for all $k \geq K$ ; and $x_{k} \in R$ with $x_{k} \notin S = R ∖ {p}$ forces $x_{k} = p$ . So $(x_{k})$ is eventually constant with value $p$ .

step 2.2A2L6

4.1

An eventually constant sequence with eventual value $p$ converges to $p$ in every topology on $R$ , since every neighbourhood of $p$ contains $p$ ; in particular $i d (x_{k}) = x_{k} \to p = i d (p)$ in the usual topology. As $(x_{k})$ and $p$ were arbitrary, $i d$ is sequentially continuous.

step 3.2A2L6

5.1

By steps 3.1 and 4.1 the map $i d : (R, T_{c o c}) \to (R, T_{R})$ is sequentially continuous and is not continuous, so the claim is false.

step 3.1step 4.1∎

Remarks

What the witness shows about the source. Assuming the Axiom of Countable Choice, the cocountable topology on $R$ is not first countable: were it, Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there would make the map above continuous, and it is not. Under that assumption some point therefore has no at most countable neighbourhood base. The sharper statement that no point has one is true but is not established here, since the refutation exhibits a single failure of continuity rather than one at every point. This is also why the cocountable topology is the standard source of examples in which sequences fail to see the topology.
The failure is not about the target. The target here is metrizable, hence as well behaved as a space can be; all the pathology is in the source, which is where sequential continuity is tested.
The same witness, worked as a counterexample with the cocountable topology's convergent sequences identified once and for all, is on the companion page (The identity from the cocountable topology on $R$ to the usual topology is sequentially continuous and not continuous ↗, In the cocountable topology on $R$ the closed sets are the countable sets and $R$ , and a sequence converges iff it is eventually constant ↗). The refutation above repeats the eventually-constant argument inline because an item on this page may not depend on an item that lives only on an examples page.

Depends on

Used by

The identity from the cocountable topology on ℝ to the usual topology is sequentially continuous and not continuous Counterexample

Dependency tree · two levels

77 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sequential space (Wikipedia) (standard reference, not scraped)
Cocountable topology (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)