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

  • $\begingroup$ @StefanH: No, I don't know how to work with uniform topologies. I do know its definition but don't know how to apply it in this case $\endgroup$ Oct 13, 2013 at 18:59
  • $\begingroup$ As Brian wrote, you should first show that $(0,0,...)$ is the only possible limit of all those sequences. To do this, take a tuple $y=(y_1,y_2,...)$ with at least one positive coordinate, say $y_k=ϵ>0$. Then see what happens if you put a neighborhood $U$ around $y$ such that pr$_k(U)\not∋0$. Can you show that there are arbitrarily large $a_n$ that are not in $U$?. This would tell you that no sequence converges to $y\ne0$ in the box topology. Could it then converge to $y$ in one of the other topologies? $\endgroup$ Oct 13, 2013 at 19:11

1 Answer 1


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.

  • $\begingroup$ Could you please work out an example for one of the sequences including box and uniform topology as I still don't know where to begin $\endgroup$ Oct 13, 2013 at 19:49
  • $\begingroup$ @RutherfordMark: Between Stefan’s comment and my hints, you have a pretty complete road map; if they’re not enough to get you started, I suspect that you don’t understand the uniform and box topologies. Can you tell me what a basic open nbhd of the zero sequence looks like in the uniform topology? $\endgroup$ Oct 13, 2013 at 19:56
  • $\begingroup$ Here for uniform topology we need to show that $d(x_n,0)<\epsilon$ That would give the neighborhood around 0. I do admit that I not very comfortable with the concept of uniform and box topology $\endgroup$ Oct 13, 2013 at 20:12
  • $\begingroup$ @RutherfordMark: Or more simply, $\langle u_n:n\in\Bbb Z^+\rangle$ is in the $\epsilon$-ball around the zero sequence iff $|u_n|<\epsilon$ for every $n\in\Bbb Z^+$. Can you see that for each $\epsilon>0$ there is an $m\in\Bbb Z^+$ such that $y_n$ is in the $\epsilon$-ball around the zero sequence whenever $n\ge m$? The same is true for the $x$ and $z$ sequences, but not for the $w$ sequence. $\endgroup$ Oct 13, 2013 at 20:18
  • $\begingroup$ But $y_n$ depends on n. and after that the sequence just has 0 so there would essentially be a "gap" of some sort $\endgroup$ Oct 13, 2013 at 20:25

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