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$\mathbb{R}^{\mathbb{N}}$ in the box topology is disconnected, the bounded and the unbounded sequences forming a separation, although every factor is connected and the product topology is connected

Statement refuted

Refuted: that a product of connected spaces is connected in the box topology. A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice proves this for the product topology only, and the restriction is not a matter of convenience.

Witness. Let $\mathbb{R}^{\mathbb{N}} := \prod_{n \in \mathbb{N}} \mathbb{R}$ , each factor carrying the usual topology (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 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), and give it the box topology $\mathcal{T}^{\square}$ (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). A point of $\mathbb{R}^{\mathbb{N}}$ is a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Put

$B \;:=\; \{\, x \in \mathbb{R}^{\mathbb{N}} : x \text{ is bounded} \,\}, \qquad V \;:=\; \{\, x \in \mathbb{R}^{\mathbb{N}} : x \text{ is unbounded} \,\}$

(Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set). Then $(B, V)$ is a separation of $\mathbb{R}^{\mathbb{N}}$ in the box topology (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets), while every factor $\mathbb{R}$ is connected and $\mathbb{R}^{\mathbb{N}}$ is connected in the product topology (A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice).

Facts & Assumptions

Given: $\mathbb{R}^{\mathbb{N}} = \prod_{n \in \mathbb{N}} \mathbb{R}$ with the box topology, and the sets $B$ and $V$ above.

[A1]

The boxes $\prod_{n} U_n$ with every $U_n$ open form a basis of the box topology; the box topology is finer than the product topology and the two differ in general (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The box topology is finer than the product topology, the two agree for a finite index set in ZF, and, assuming the Axiom of Choice for nonempty factors, the box topology is strictly finer whenever infinitely many factors have a nonempty proper open subset, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A2]

A sequence of reals $x$ is bounded when there is $M \in \mathbb{R}$ with $|x_n| \le M$ for every $n \in \mathbb{N}$ , and unbounded otherwise (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set).

[A3]

$(a,b) = \{\, t : a < t < b \,\}$ is open in $\mathbb{R}$ , and $|s - t| < 1$ gives $|s| \le |t| + 1$ and $|s| \ge |t| - 1$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, 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 metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[A4]

A separation of a space is a pair of open, nonempty, disjoint sets covering it; a space admitting one is disconnected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, For a topological space the following agree: no separation exists, the only clopen subsets are $\varnothing$ and $X$ , and every continuous map to the two-point discrete space is constant, claim 1).

[A5]

$\mathbb{R}$ is connected, being order-convex, and a product of connected spaces is connected in the product topology (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$ ", A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[A6]

For every real $\varepsilon > 0$ there is a natural $k \ge 1$ with $1/k < \varepsilon$ ; the canonical naturals of $\mathbb{R}$ are unbounded above (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Counterexample

technique · direct

1.1

$B$ and $V$ are disjoint and cover $\mathbb{R}^{\mathbb{N}}$ , a sequence being bounded or unbounded and not both, by [A2].

1.2

Both are nonempty: the constant sequence $x_n := 0$ is bounded by $M = 0$ , and the sequence $y_n := \iota(n)$ of canonical naturals is unbounded by [A6], no real bounding all of them.

A2A6

2.1

$B$ is open in the box topology. Let $x \in B$ with bound $M$ , and let $W := \prod_{n} (x_n - 1,\ x_n + 1)$ , a box, hence open by [A1]. For $y \in W$ one has $|y_n - x_n| < 1$ , so $|y_n| \le |x_n| + 1 \le M + 1$ for every $n$ by [A3]; hence $y \in B$ and $W \subseteq B$ .

step 1.1A1A2A3

2.2

$V$ is open in the box topology. Let $x \in V$ and take the same box $W := \prod_{n} (x_n - 1,\ x_n + 1)$ , open by [A1]. For $y \in W$ and any $M \in \mathbb{R}$ , unboundedness of $x$ gives $n$ with $|x_n| > M + 1$ , and then $|y_n| \ge |x_n| - 1 > M$ by [A3]; so no $M$ bounds $y$ , that is $y \in V$ and $W \subseteq V$ .

step 1.1A1A2A3

3.1

By steps 1.1, 1.2, 2.1 and 2.2 the pair $(B,V)$ is a separation of $\mathbb{R}^{\mathbb{N}}$ in the box topology, so that space is disconnected by [A4]; whereas every factor is connected and the same product is connected in the product topology by [A5].

step 1.1step 1.2step 2.1step 2.2A4A5∎

Remarks

The proof of the product-topology theorem breaks at exactly one place. There the finite-support points around a base point are dense, because a basic product-open set constrains only finitely many coordinates. A box constrains every coordinate at once, so a point differing from the base point in finitely many coordinates need not lie in a given box, the density argument fails, and with it the conclusion.
The separating property is invariant under bounded perturbation, and that is all that is needed. Boundedness of a sequence is unchanged by moving every coordinate by less than $1$ , and a single box of width $2$ around a point performs exactly such a perturbation. Any property with that stability separates the box topology in the same way.
Nothing here contradicts The box topology is finer than the product topology, the two agree for a finite index set in ZF, and, assuming the Axiom of Choice for nonempty factors, the box topology is strictly finer whenever infinitely many factors have a nonempty proper open subset. The box topology is finer, so it has more open sets and therefore more chances to separate; a finer topology can disconnect a space that a coarser one connects, and this witness is that phenomenon in its simplest form.

Depends on

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Sources

Box topology (Wikipedia) (standard reference, not scraped)
Connected space (Wikipedia) (standard reference, not scraped)

RN\mathbb{R}^{\mathbb{N}}RN in the box topology is disconnected, the bounded and the unbounded sequences forming a separation, although every factor is connected and the product topology is connected