Editing Box topology (section)

=== 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.