Pre-image of continuous functions I am trying to solve the following problem:
In which topologies (uniform, product, box) are the following functions from $\mathbb{R}$ to $\mathbb{R^\omega}$ continuous? 
$f(t)=(t,2t,3t,....)$
$g(t)=(t,t,t,.......)$
$h(t)= (t,\frac{1}{2}t, \frac{1}{3}t,...)$
I am having trouble with the box topology and uniform topology. In my solutions it says for the box topology consider, $B=(-1,1)\times (\frac{-1}{4}.\frac{1}{4})\times\ldots$ has pre-image of {0} for all the functions. But how do we know this? Doesn't this depend on different values of $t$ though for each function?
For uniform topology, is there a nice way to construct the open balls?
( I have asked a similar question before. Based on my understanding from the answers over there I am trying to solve this problem and am stuck!)
 A: The fact that $B=(-1,1)\times(-\frac14,\frac14)\times \ldots\times (-\frac1{n^2},\frac1{n^2})\}\times\ldots$ is an open set in the box topology should be clear.
We have $$\begin{eqnarray}f(t)\in B&\iff &\forall n\in\omega\colon nt\in(-\frac1{n^2},\frac1{n^2})\iff\forall n\in\omega \colon t\in(-\frac1{n^3},\frac1{n^3})\\
g(t)\in B&\iff& \forall n\in\omega\colon nt\in(-\frac1{n^2},\frac1{n^2})\\
h(t)\in B&\iff &\forall n\in\omega\colon \frac1nt\in(-\frac1{n^2},\frac1{n^2})\iff\forall n\in\omega \colon t\in(-\frac1{n},\frac1{n})\\\end{eqnarray}$$
Since already$$\bigcap_{n\in\omega}(-\frac1n,\frac1n)=\{0\} $$
the claim follows and for all these functions the preimage of the opeb set $B$ is the non-open set $\{0\}$.
For uniform topology it is best not to think of open balls, but use convergence directly (after all that is the definition).
A: Let $t \in \mathbb R$, $t \ne 0$. By the Archimedean property, we can find $n \in \mathbb N$ such that $0 < 1/n^2 < |t|$. But this means that the $n^\text{th}$ component of $f(t)$ (which is $t$) is not in the $n^\text{th}$ component of $B$ (which is $(-1/n^2, 1/n^2)$). Therefore $t$ cannot be in the pre-image of $B$. We conclude that the pre-image of $B$ is $\{0\}$ and $f$ is not continuous.
$g$ and $h$ can be handled similarly.
