Define $f: \mathbb R \to \mathbb R^\omega$ such that $f(t) = (t, 2t, 3t, \dots)$, where $\mathbb R$ is given the standard topology, and $\mathbb R^\omega$ is given the uniform topology. I'm trying to prove f is not continuous. Here is my attempt at a counterexample;
Let $U = B(0,1) = \{x \in \mathbb R^\omega: ||x|| < 1\}$. Now $U$ is open as an open unit ball, and $f^{-1}(U)$ is such that if $x \in f^{-1}(U)$ then $nx < 1$ for all $n \in \mathbb Z_+$. In particular this implies $x < \frac{1}{n}$ for all $n$ so $x = 0$ and $f^{-1}(U) = \{0\}$ is closed. Is this a correct counterexample?
Why does $f^{-1}(U)$ is closed mean $f$ is not continuous? (Yes, I know how it would go, but it is a rather long-winded way to do so.)
It is better to say $f^{-1}(U)=\{0\}$ is not open, then it is just the definition of continuity to reach the desired conclusion.