Box topology

The following example is based on the Hilbert cube. Let R^ω denote the countable cartesian product of R with itself, i.e. the set of all sequences in R. Equip R with the standard topology and R^ω with the box topology. Define:

${\displaystyle {\begin{cases}f:\mathbf {R} \to \mathbf {R} ^{\omega }\\x\mapsto (x,x,x,\ldots )\end{cases}}}$

So all the component functions are the identity and hence continuous, however we will show f is not continuous. To see this, consider the open set

${\displaystyle U=\prod _{n=1}^{\infty }\left(-{\tfrac {1}{n}},{\tfrac {1}{n}}\right).}$

Suppose f were continuous. Then, since:

${\displaystyle f(0)=(0,0,0,\ldots )\in U,}$

there should exist ${\displaystyle \varepsilon >0}$ such that ${\displaystyle (-\varepsilon ,\varepsilon )\subset f^{-1}(U).}$ But this would imply that

${\displaystyle f\left({\tfrac {\varepsilon }{2}}\right)=\left({\tfrac {\varepsilon }{2}},{\tfrac {\varepsilon }{2}},{\tfrac {\varepsilon }{2}},\ldots \right)\in U,}$

which is false since ${\displaystyle {\tfrac {\varepsilon }{2}}>{\tfrac {1}{n}}}$ for ${\displaystyle n>{\tfrac {2}{\varepsilon }}.}$ Thus f is not continuous even though all its component functions are.

Consider the countable product ${\displaystyle X=\prod X_{i}}$ where for each i, ${\displaystyle X_{i}=\{0,1\}}$ with the discrete topology. The box topology on ${\displaystyle X}$ will also be the discrete topology. Since discrete spaces are compact if and only if they are finite, we immediately see that ${\displaystyle X}$ is not compact, even though its component spaces are.

${\displaystyle X}$ is not sequentially compact either: consider the sequence ${\displaystyle \{x_{n}\}_{n=1}^{\infty }}$ given by

${\displaystyle (x_{n})_{m}={\begin{cases}0&m<n\\1&m\geq n\end{cases}}}$

Since no two points in the sequence are the same, the sequence has no limit point, and therefore ${\displaystyle X}$ is not sequentially compact.

Topologies are often best understood by describing how sequences converge. In general, a Cartesian product of a space ${\displaystyle X}$ with itself over an indexing set ${\displaystyle S}$ is precisely the space of functions from ${\displaystyle S}$ to ${\displaystyle X}$, denoted ${\textstyle \prod _{s\in S}X=X^{S}}$. The product topology yields the topology of pointwise convergence; sequences of functions converge if and only if they converge at every point of ${\displaystyle S}$.

Because the box topology is finer than the product topology, convergence of a sequence in the box topology is a more stringent condition. Assuming ${\displaystyle X}$ is Hausdorff, a sequence ${\displaystyle (f_{n})_{n}}$ of functions in ${\displaystyle X^{S}}$ converges in the box topology to a function ${\displaystyle f\in X^{S}}$ if and only if it converges pointwise to ${\displaystyle f}$ and there is a finite subset ${\displaystyle S_{0}\subset S}$ and there is an ${\displaystyle N}$ such that for all ${\displaystyle n>N}$ the sequence ${\displaystyle (f_{n}(s))_{n}}$ in ${\displaystyle X}$ is constant for all ${\displaystyle s\in S\setminus S_{0}}$. In other words, the sequence ${\displaystyle (f_{n}(s))_{n}}$ is eventually constant for nearly all ${\displaystyle s}$ and in a uniform way.[3]