Box topology and a non-continuous function

Question

I'm having trouble understanding box topologies on infinite spaces.

I'm asked to prove that the function f going from ([0,1], standard top.) to ([0,1]^N, box top.) is not continuous. This function is defined as f(x) = (x, x, x, ...)

To do so, I must show that there exists an open set in the box topology such that its preimage is closed in [0,1]. But my question is, if U = open in box topology, then every member of U has a little space to wiggle and stay within U, so wouldn't that imply that each component can also wiggle a little bit, implying that the preimage of U has to be open?

Answer 1

No, you must show that there is an open set $O$ in the box topology so that $f^{-1}[O]$ is not open (not the same as closed!).

Try $$O= \prod_{n =1}^\infty (-\frac1n,\frac1n)$$ and show that $f^{-1}[O]=\{0\}$, indeed not open.

Answer 2

This is, in fact, a standard example to show that box topology does not have that nice properties as product topology.

We want to find an open set $U \subseteq [0,1]^\mathbb N$ (with box topology) such that $f^{-1}U \subseteq [0,1]$ (with standard topology) is not open. What if we put e.g. $U := [0,1) \times [0, \frac12) \times \dots \times [0,\frac1n) \times \dots$?