Is $\ell^\infty$ with box topology connected?

Question

Let $X = \ell^\infty$ be all bounded real sequences and equip $X$ with subspace topology $X\subseteq \square_{n=1}^\infty \mathbb{R}$ where $\square_{n=1}^\infty \mathbb{R}$ is box product of countable amount of copies of $\mathbb{R}$, that is the basis for its topology is $\prod_{n=1}^\infty U_n$ where $U_n\subseteq \mathbb{R}$ are open.

Sets of the form $a+X$ for $a\in\square_{n=1}^\infty \mathbb{R}$ partition $\square_{n=1}^\infty \mathbb{R}$ into clopen homeomorphic subsets, so it makes sense to study $X$ instead of $\square_{n=1}^\infty \mathbb{R}$.

Let $f:[0, 1]\to X$ be $f(t) = tx+(1-t)y$ where $x, y$ differ by infinite amount of terms. Then $f$ is not continuous: assume without loss of generality that $y = 0$ and let $U_n = \mathbb{R}$ if $x_n = 0$ and $(-|x_n|/n, |x_n|/n)$ otherwise. Then $f(0) = 0\in \prod_n U_n$ but $f(t)\notin \prod_n U_n$ for $t > 0$. So if $X$ is path-connected, its not obvious.

Consider $A(x) = \{y\in X : y_n = x_n\text{ for all but finite amount of }n\}$. Then $A(x)$ is path-connected.

Is $X$ a connected space?

Answer 1

It is not connected. Let $C=\{(x_n)\in l^\infty\ |\ x_n\to 0\}$. It is clopen.

Indeed, for a given sequence $(x_n)$ consider $U=\prod (x_n-1/n,x_n+1/n)$. If $x_n$ converges to $0$ then so does every sequence in $U$. And so $C$ is open. And vice versa, if a $x_n$ does not converge to $0$, then no sequence in $U$ converges to $0$. And so $C$ is closed.

Answer 2

I'll prove that distinct $A(x) = \{y\in \square_{n=1}^\infty \mathbb{R} : x_n = y_n\text{ for all but finitely many }n\}$ are quasi-components of $\square_{n=1}^\infty \mathbb{R}$, that is $A(x)$ is the intersection of all clopen sets containing $x$. Since $A(x)$ is path-connected, it follows they are also connected components and path-components of $\square_{n=1}^\infty \mathbb{R}$.

Observe that if $Z = \{x\in \square_{n=1}^\infty \mathbb{R} : x_n\to 0\}$ then $Z$ is clopen in $\square_{n=1}^\infty \mathbb{R}$ (see @freakish answer), (non-topological) subgroup, and sets of the form $a+Z$ with $a\in \square_{n=1}^\infty \mathbb{R}$ partition $\square_{n=1}^\infty \mathbb{R}$ into clopen subsets, and moreover $a+A(x) = A(x+a)$.

So its enough to show that $A(x)$ for $x\in Z$ are quasi-components of $Z$.

If $z\notin A(x)$ then we need to show there is a clopen set $C$ with $z\in C$ and $x\notin C$. By translating we can assume $z = 0$.

The idea is that we can replace $x_n\to 0$ in freakish answer, by $x_n$ convering to $0$ appropriately fast.

Let $$C = \{y\in Z : \exists_M \forall_n (x_n\neq 0\implies |y_n|\leq M|x_n|^2)\}.$$

Given $y\in Z$ define $U = \prod_n U_n\cap Z$ where $U_n = (-|x_n|^2+y_n, |x_n|^2+y_n)$ if $x_n\neq 0$ and $\mathbb{R}$ otherwise.

Suppose $y\in C$. If $z\in U$ and $x_n\neq 0$ then $|z_n| \leq |x_n|^2+|y_n| \leq (M+1)|x_n|^2$ so that $z\in C$. So $C$ is open.

Suppose $y\notin C$. If $z\in U$ and $M\geq 0$ then there is $n$ with $x_n\neq 0$ but $|y_n| > M|x_n|^2$ so $|z_n| \geq |y_n|-|z_n-y_n| > (M-1)|x_n|^2$. So $C$ is closed.

Its clear that $0\in C$. But $x\notin C$ for otherwise we would find $M$ such that $|x_n|\leq M|x_n|^2$ for all $n$, and so $1\leq M|x_n|$ whenever $x_n\neq 0$, which is impossible since $x_n\to 0$ and $x_n\neq 0$ for infinitely many $n$.