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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

Example

Give $\mathbb{R}$ the cocountable topology $\mathcal{T}_{\mathrm{coc}}$ , whose open sets are $\varnothing$ 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). Then:

The closed sets are exactly the at most countable subsets of $\mathbb{R}$ together with $\mathbb{R}$ itself, and these two families are disjoint, $\mathbb{R}$ being uncountable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)). In particular every singleton is closed.
Closures. For $A \subseteq \mathbb{R}$ , $\overline{A} = \begin{cases} A & A \text{ at most countable} \\ \mathbb{R} & A \text{ uncountable.} \end{cases}$
A sequence converges if and only if it is eventually constant (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), and then it converges to its eventual value and to no other point.

Claim 3 is what makes this space the standard witness that sequences can be blind to a topology: the convergent sequences are the same as in the discrete topology, while the topology itself is very far from discrete by claim 2.

Facts & Assumptions

Given: $\mathbb{R}$ with the cocountable topology, a subset $A \subseteq \mathbb{R}$ , a sequence $(x_k)$ in $\mathbb{R}$ and points $p, q \in \mathbb{R}$ . Write $R := \{\, x_k : k \in \mathbb{N} \,\}$ for the range of $(x_k)$ .

[A1]

The open sets of $\mathcal{T}_{\mathrm{coc}}$ are $\varnothing$ together with the sets of at most countable complement; a set is closed exactly when its complement is open (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A2]

$x_k \to p$ means that for every neighbourhood $N$ of $p$ there is $K$ with $x_k \in N$ for all $k \ge K$ ; a neighbourhood of $p$ is a set containing an open set containing $p$ , 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).

[L1]

$\mathbb{R}$ is uncountable, that is not at most countable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874), Finite, countably infinite, countable, uncountable).

[L2]

Every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[L3]

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

[L4]

$\overline{A}$ is the smallest closed superset of $A$ , and $A$ is closed exactly when $A = \overline{A}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2).

Verification

technique · direct

1.1

A set $F \subseteq \mathbb{R}$ is closed exactly when $\mathbb{R} \setminus F$ is open, that is exactly when $\mathbb{R} \setminus F = \varnothing$ , giving $F = \mathbb{R}$ , or $\mathbb{R} \setminus (\mathbb{R} \setminus F) = F$ is at most countable. So the closed sets are $\mathbb{R}$ together with the at most countable sets, and $\mathbb{R}$ is not among the latter by [L1].

A1L1

1.2

A singleton is finite, hence at most countable, hence closed.

A1L1

1.3

Assume $x_k \to p$ . The map $k \mapsto x_k$ is a surjection $\mathbb{N} \to R$ and $R \ne \varnothing$ , so $R$ is at most countable by [L3]; hence $S := R \setminus \{p\}$ is at most countable by [L2], and $U := \mathbb{R} \setminus S$ is open by [A1] and contains $p$ .

assume-hypA1L2L3

1.4

Conversely, if $(x_k)$ is eventually constant with value $q$ , say $x_k = q$ for all $k \ge K_0$ , then for every neighbourhood $N$ of $q$ one has $q \in N$ and hence $x_k \in N$ for all $k \ge K_0$ ; so $x_k \to q$ .

2.1

If $A$ is at most countable then $A$ is closed by step 1.1, so $\overline{A} = A$ by [L4].

step 1.1L4

2.2

If $A$ is uncountable then no at most countable set contains $A$ , since a subset of an at most countable set is at most countable by [L2]; so the only closed superset of $A$ is $\mathbb{R}$ and $\overline{A} = \mathbb{R}$ .

step 1.1L2L4

2.3

By [A2] applied to the neighbourhood $U$ of step 1.3 there is $K$ with $x_k \in U$ for all $k \ge K$ ; and $x_k \in R$ together with $x_k \notin S = R \setminus \{p\}$ forces $x_k = p$ . So $(x_k)$ is eventually constant with value $p$ .

step 1.3A2

2.4

Suppose $(x_k)$ is eventually constant with value $q$ , say $x_k = q$ for all $k \ge K_0$ , and let $p \ne q$ . The set $N := \mathbb{R} \setminus \{q\}$ is open by [A1], its complement $\{q\}$ being finite, and $p \in N$ , so $N$ is a neighbourhood of $p$ ; but $x_k = q \notin N$ for every $k \ge K_0$ , so no tail of the sequence lies in $N$ and $x_k \not\to p$ . Hence the eventual value is the only limit.

step 1.4A1A2

3.1

Claim 1 is step 1.1 with step 1.2, claim 2 is steps 2.1 and 2.2, and claim 3 is steps 2.3, 1.4 and 2.4.

step 1.1step 1.2step 2.1step 2.2step 2.3step 1.4step 2.4∎

Remarks

Assuming the Axiom of Countable Choice, this space is not first countable. Under that hypothesis Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there would otherwise force $\operatorname{seqcl}(A)=\overline{A}$ for every $A$ , whereas claim 3 makes the sequential closure of $[0,1]$ equal to $[0,1]$ and claim 2 makes its closure all of $\mathbb{R}$ (In the cocountable topology on $\mathbb{R}$ the sequential closure of $[0,1]$ is $[0,1]$ while its closure is all of $\mathbb{R}$ ). Under the same hypothesis it is therefore not metrizable either.
No two nonempty open sets are disjoint here either. If $U$ and $V$ are nonempty and open then $\mathbb{R} \setminus (U \cap V)$ is a union of two at most countable sets and hence at most countable, so $U \cap V$ cannot be empty, $\mathbb{R}$ being uncountable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)). The argument is the one used for the cofinite topology (On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint) with "at most countable" in place of "finite".
The topology is strictly coarser than the discrete topology and strictly finer than the cofinite topology on $\mathbb{R}$ . It is strictly coarser than discrete because $\{0\}$ is not open, its complement being uncountable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874), Every subset of an at most countable set is at most countable). It is finer than cofinite because a finite set is at most countable, and strictly so because $\mathbb{R} \setminus \mathbb{N}$ has an at most countable complement that is not finite.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 18 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

Cocountable topology (Wikipedia) (standard reference, not scraped)
Countable set (Wikipedia) (standard reference, not scraped)

In the cocountable topology on R\mathbb{R}R the closed sets are the countable sets and R\mathbb{R}R, and a sequence converges iff it is eventually constant