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The identity from the cocountable topology on $\mathbb{R}$ to the usual topology is sequentially continuous and not continuous

Statement refuted

Refuted: that a sequentially continuous map of topological spaces is continuous (FALSE: a sequentially continuous map between topological spaces is continuous).

Witness. Let $\mathcal{T}_{\mathrm{coc}}$ be the cocountable topology on $\mathbb{R}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$ , and a sequence converges iff it is eventually constant) and $\mathcal{T}_{\mathbb{R}}$ its usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $\mathbb{R}$ 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 identity $\mathrm{id} : (\mathbb{R}, \mathcal{T}_{\mathrm{coc}}) \longrightarrow (\mathbb{R}, \mathcal{T}_{\mathbb{R}})$ is sequentially continuous (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure) and is not continuous (Continuity of a map of topological spaces at a point and globally).

This is the witness inlined in the refutation of FALSE: a sequentially continuous map between topological spaces is continuous, recorded here with the convergent sequences of the source identified once and for all in In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$ , and a sequence converges iff it is eventually constant rather than re-derived.

Facts & Assumptions

Given: $\mathbb{R}$ carrying $\mathcal{T}_{\mathrm{coc}}$ as source and $\mathcal{T}_{\mathbb{R}}$ as target, and the identity function between them.

[A1]

The open sets of $\mathcal{T}_{\mathrm{coc}}$ are $\varnothing$ together with the sets of at most countable complement (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L1]

In $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ a sequence converges if and only if it is eventually constant, and then to its eventual value (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$ , and a sequence converges iff it is eventually constant, claim 3).

[L2]

$B(x,r) = (x-r,\ x+r)$ in the usual topology, and every ball is open there (The absolute value makes $\mathbb{R}$ 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, Open ball, closed ball and sphere in a metric space, 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, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

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)); sequential continuity at $p$ says that $x_k \to p$ implies $f(x_k) \to f(p)$ , and every point lies in each of its neighbourhoods (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L4]

For $a < b$ the interval $(a,b)$ is uncountable (Every nondegenerate interval of $\mathbb{R}$ is uncountable), and every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L5]

$0 < 1$ and hence $1 < 1+1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities).

Counterexample

technique · direct

1.1

$V := B(0,1) = (-1,1)$ is open in the usual topology, the radius $1$ being positive by [L5].

L2L5

1.2

$1 < 1+1$ by [L5], so $(1,\ 1+1)$ is uncountable by [L4], and it is contained in $\mathbb{R} \setminus (-1,1)$ , a point $x > 1$ satisfying neither $x < 1$ nor $-1 < x < 1$ .

L4L5

1.3

Let $(x_k)$ be a sequence converging to $p$ in $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ ; by [L1] it is eventually constant with value $p$ , say $x_k = p$ for all $k \ge K$ .

2.1

$\mathbb{R} \setminus (-1,1)$ is not at most countable, since otherwise its subset $(1,\ 1+1)$ would be at most countable by [L4], contradicting step 1.2. Hence $V = (-1,1)$ is nonempty and has a complement that is not at most countable, so $V \notin \mathcal{T}_{\mathrm{coc}}$ .

step 1.2A1L4

2.2

The image sequence $\mathrm{id}(x_k) = x_k$ of step 1.3 is eventually equal to $p$ , so for every neighbourhood $N$ of $p$ in the usual topology one has $p \in N$ and hence $x_k \in N$ for all $k \ge K$ ; that is $\mathrm{id}(x_k) \to \mathrm{id}(p)$ in the usual topology. As $(x_k)$ and $p$ were arbitrary, $\mathrm{id}$ is sequentially continuous.

step 1.3L3

3.1

$\mathrm{id}^{-1}[V] = V$ is open in the target by step 1.1 and not open in the source by step 2.1, so $\mathrm{id}$ is not continuous; with step 2.2 the witness is established and the claim of FALSE: a sequentially continuous map between topological spaces is continuous is refuted.

step 1.1step 2.1step 2.2L3∎

Remarks

The two topologies are incomparable, so the identity is discontinuous in both directions. The set $(-1,1)$ is open in the usual topology and not in the cocountable one, which is the counterexample above. In the other direction $\mathbb{R} \setminus \mathbb{Q}$ is cocountable-open, $\mathbb{Q}$ being at most countable ( $\mathbb{Q}$ is countably infinite), and is not open in the usual topology, since every ball contains a rational (The rationals embed densely in the reals). So neither topology is finer than the other, and this pair is not an instance of the continuous-bijection failure recorded in FALSE: every continuous bijection of topological spaces is a homeomorphism, which needs two comparable topologies.
Why sequences are blind here. A sequence visits at most countably many points, and the cocountable topology supplies a neighbourhood of its proposed limit omitting every other point in that range; this is exactly the mechanism used in step 1.3 through [L1]. Assuming the Axiom of Countable Choice, Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there then shows that this failure of the sequential test forces the source not to be first countable.
The target's good behaviour is irrelevant. It is metrizable, hence as well behaved as possible, and the failure is entirely on the source side, which is where sequential continuity is tested.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 108 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

The identity from the cocountable topology on R\mathbb{R}R to the usual topology is sequentially continuous and not continuous