Editing Box topology (section)

==Properties==

Box topology on '''R'''<sup>''ω''</sup>:<ref>Steen, Seebach, 109. pp. 128&ndash;129.</ref>

* The box topology is [[completely regular]]
* The box topology is neither [[compact space|compact]] nor [[Connected space|connected]]
* The box topology is not [[first countable]] (hence not [[metrizable]])
* The box topology is not [[separable space|separable]]
* The box topology is [[paracompact]] (and hence normal and completely regular) if the [[continuum hypothesis]] is true
 
=== Example — failure of continuity ===
The following example is based on the [[Hilbert cube]]. Let '''R'''<sup>''ω''</sup> denote the countable cartesian product of '''R''' with itself, i.e. the set of all [[sequence]]s in '''R'''. Equip '''R''' with the [[Real line#As a topological space|standard topology]] and '''R'''<sup>''ω''</sup> with the box topology. Define:

:<math>\begin{cases} f : \mathbf{R} \to \mathbf{R}^\omega \\ x \mapsto (x,x,x, \ldots) \end{cases}</math>

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 

:<math> U = \prod_{n=1}^{\infty} \left ( -\tfrac{1}{n}, \tfrac{1}{n} \right ).</math>

Suppose ''f'' were continuous. Then, since:

:<math>f(0) = (0,0,0, \ldots ) \in U,</math>

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

:<math> f\left (\tfrac{\varepsilon}{2} \right ) = \left ( \tfrac{\varepsilon}{2}, \tfrac{\varepsilon}{2}, \tfrac{\varepsilon}{2}, \ldots \right ) \in U,</math>

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

=== Example — failure of compactness ===
Consider the countable product <math>X = \prod_{i \in \N} X_i</math> where for each ''i'', <math>X_i = \{0,1\}</math> with the discrete topology. The box topology on <math>X</math> will also be the discrete topology. Since discrete spaces are compact if and only if they are finite, we immediately see that <math>X</math> is not compact, even though its component spaces are. 

<math>X</math> is not sequentially compact either: consider the sequence <math>\{x_n\}_{n=1}^\infty</math> given by 
:<math>(x_n)_m=\begin{cases}
  0 & m < n \\
  1 & m \ge n 
\end{cases}</math>
Since no two points in the sequence are the same, the sequence has no limit point, and therefore <math>X</math> is not sequentially compact.

===Convergence in the box topology===
Topologies are often best understood by describing how sequences converge. In general, a Cartesian product of a space <math>X</math> with itself over an [[index set|indexing set]] <math>S</math> is precisely the space of functions from <math>S</math> to <math>X</math>'','' denoted <math display="inline">\prod_{s \in S} X = X^S</math>. The product topology yields the topology of [[pointwise convergence]]; sequences of functions converge if and only if they converge at every point of <math>S</math>.

Because the box topology is finer than the product topology, convergence of a sequence in the box topology is a more stringent condition.  Assuming <math>X</math> is Hausdorff, a sequence <math>(f_n)_n</math> of functions in <math>X^S</math> converges in the box topology to a function <math>f\in X^S</math> if and only if it converges pointwise to <math>f</math> and 
there is a finite subset <math>S_0\subset S</math> and there is an <math>N</math> such that for all <math>n>N</math> the sequence <math>(f_n(s))_n</math> in <math>X</math> is constant for all <math>s\in S\setminus S_0</math>.  In other words, the sequence <math>(f_n(s))_n</math> is eventually constant for nearly all <math>s</math> and in a uniform way.<ref>{{cite web|last1=Scott|first1=Brian M.|title=Difference between the behavior of a sequence and a function in product and box topology on same set|url=https://math.stackexchange.com/q/448575|website=math.stackexchange.com}}</ref>