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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.


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@Srivatsan: I suspect that the first argument is missing a pair of parentheses: $B_\rho\left(\langle 1-\frac1{2^n}:n\in\mathbb{N}\rangle,1\right)$. – Brian M. Scott Oct 4 '11 at 22:01
@DavideGiraudo, done. – Chulumba Oct 4 '11 at 22:03
@Srivatsan, the expression inside the bracket is the open ball with radius $1$ centered at the $(1-\frac{1}{2^n})_{n\in \mathbb{N}}$. – Chulumba Oct 4 '11 at 22:03
If $t>1$, $g(t)$ will be in the ball iff $|t-1/2|<1$, so ... – Brian M. Scott Oct 4 '11 at 22:17
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 $1$. – Srivatsan Oct 4 '11 at 22:24
up vote 1 down vote accepted

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 $1<t<\frac32$; what can you say about $$\min\left\{\left|t-\frac1{2^n}\right|,1\right\}$$ for arbitrary $n\in\mathbb{N}$?

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