# Convergence of sequence in uniform and box topologies

I am trying the following problem:

$w_1=(1,1,1,1,\ldots)$
$w_2=(0,2,2,\ldots)$
$w_3=(0,0,3,3,\ldots)$ $\cdots$

$x_1=(1,1,1,1,\ldots)$
$x_2=(0,\frac{1}{2},\frac{1}{2},\frac{1}{2}\ldots)$
$x_3=(0,0,\frac{1}{3},\frac{1}{3},\ldots)$ $\cdots$

$y_1=(1,0,0,0,\ldots)$
$y_2=(\frac{1}{2},\frac{1}{2},0,0,\ldots)$
$y_3=(\frac{1}{3},\frac{1}{3},\frac{1}{3},0\ldots)$ $\cdots$

$z_1=(1,1,0,0,\ldots)$
$z_2=(\frac{1}{2},\frac{1}{2},0,0,\ldots)$
$z_3=(\frac{1}{3},\frac{1}{3},0,0\ldots)$ $\cdots$

I have been able to prove that they all are convergent in product topology but have no idea how we can prove for uniform and box topologies. Any help will be appreciated

#### Solutions Collecting From Web of "Convergence of sequence in uniform and box topologies"

The last one is the easiest: in all three topologies it’s essentially a sequence in $\Bbb R^2$.

Show that the first is not convergent in the uniform topology and therefore cannot be convergent in the finer box topology. You can do this by showing that the zero sequence is the only possible limit and then showing that it is not the limit in the uniform topology.

For the second and third, show first that the only possible limit in either topology is the zero sequence. Then show that in both cases the zero sequence is the limit in the uniform topology but not in the box topology. You may want to consider the set

$$\prod_{n\in\Bbb Z^+}\left(-\frac1{n+1},\frac1{n+1}\right)\;,$$

which is open in the box topology.