Continuity in Box topology

Question

Suppose that $f:[0,1]\rightarrow [0,1]^{\mathbb{N}}$, with $f(x)=(x,x,x,x,..)$.Then is this function continious in box topology ?? Where box topogoly is generated by $\mathcal{B}=\left\{\prod_{n\in\mathbb{N}}U_{n}:U_{n} \ \text{is open in} \ [0,1]\right\}$ .

Answer 1

No, it is not. Pick a sequence of open subsets $(U_n)_{n\in\mathbb N}$ of $[0,1]$ such that $\bigcap_{n\in\mathbb N}U_n$ is not open. But $\bigcap_{n\in\mathbb N}U_n=f^{-1}\left(\prod_{n\in\mathbb N}U_n\right)$.

Answer 2

No it is not: Define $U_n := (\frac{1}{2}-\frac{1}{n+1}, \frac{1}{2}+\frac{1}{n+1})$. Then by your basis $U := \prod_{n}{U_n}$ will be an open set in $[0,1]^\mathbb{N}$. But since $\frac{1}{n+1} \to 0$ one has $f^{-1}(U) = \{\frac{1}{2}\}$ which is not open in $[0,1]$.

Answer 3

Consider the convergent sequence in $[0, 1]$, $a_n = \frac{1}{n} \to 0$. However, $f(a_n) \not\to f(0) = (0, 0, \ldots)$ as $n \to \infty$ since for all $n$, $f(a_n) \notin \prod_{k=1}^\infty [0, \frac{1}{k})$ which is a neighborhood of $(0, 0, \ldots)$ in the box topology.

Therefore, $f$ is not continuous (in particular, this shows that $f$ is not continuous at 0, and a similar argument should show $f$ is not continuous at any point).