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Consider the function $g: \mathbb{R}\to \mathbb{R}^\omega$ given by $g(t)=(t,t,t,...)$ where $ \mathbb{R}^\omega$ is in the uniform topology. Can we find the exact answer to $g^{-1}(B_{\rho}((1-\frac{1}{2^n})_{n\in \mathbb{N}}, 1))$ where $\rho$ is the uniform metric as defined in Munkres' Topology book (page ~120)) i.e. $\rho(x,y) = \sup_i\{ \min\{|x_i - y_i|, 1\}\}$

I have been able to show that the pre-image is of the form $(0,1+\epsilon)$ but I can't determine the exact value of the right hand side limit.

Regards.

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    To elaborate @Brian's answer, if t > 1, then the distance between $g(t)$ and the center of the open ball is simply: $ \min \{1, \sup_n (t+2^{-n}-1) \} . $ The expression inside the sup is maximized for $n=1$ (or whatever the starting index is). And you want this distance to be strictly smaller than $1$.2011-10-04

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You’ve done the hard part in getting $0$ as the left endpoint of but not included in the inverse image. Suppose that $t>1$ and $g(t)$ is in the ball. Then certainly $\min\left\{\left|t-\frac12\right|,1\right\}<1,$ so $1; what can you say about $\min\left\{\left|t-\frac1{2^n}\right|,1\right\}$ for arbitrary $n\in\mathbb{N}$?