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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 $\mathcal{B}=\left\{\prod_{n\in\mathbb{N}}U_{n}:U_{n} \ \text{is open in} \ [0,1]\right\}$ .

José Carlos Santos
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Jonathan1234
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  • I don't think it is, can you convince me @mathsta ? – Jürgen Sukumaran Feb 27 '18 at 20:50
  • Is the inverse image of an open set open? What have you done on this problem? – saulspatz Feb 27 '18 at 20:51
  • I'm not familiar with box topology I just proved that this function is continious with product topology because every coordinate is id mapping wich is continuous and I want to see if we can still hold continuity with box topology – Jonathan1234 Feb 27 '18 at 20:58
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    Check my response here. Your question is a special case of that question wherein we have $f:[0,1] \rightarrow [0,1]^{\mathbb{N}}$ such that $f(x) = \langle f_k(x) \rangle_{k=1}^\infty$. So each component function $f_k$, which gives the $k^\text{th}$ coordinate of the output, is, in this scenario, just the identity on $[0,1]$. – Kaj Hansen Feb 27 '18 at 21:34

3 Answers3

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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 $\bigcap_{n\in\mathbb N}U_n=f^{-1}\left(\prod_{n\in\mathbb N}U_n\right)$.

José Carlos Santos
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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 $[0,1]$.

lush
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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).

Daniel Schepler
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