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Let $\mathbb{R}^\omega$ denote the set of all sequences of real numbers, let $\tilde{\rho}$ denote the uniform metric on $\mathbb{R}^\omega$ defined as $$ \begin{align} \tilde{\rho}(x,y) & \colon= \sup \left\{ \ \min \{ \ \vert x_n - y_n \vert, 1 \} \ \colon \ n = 1, 2, 3, \ldots \ \right\} \\ & \mbox{ for all } \ x \colon= (x_n)_{n \in \mathbb{N}}, \ y \colon= (y_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega. \end{align} $$

Let $0 < \epsilon < 1$, and let $$ U(x,\epsilon) \ \colon= \ (x_1 - \epsilon, x_1 + \epsilon) \times (x_2 - \epsilon, x_2 + \epsilon) \times (x_3 - \epsilon, x_3 + \epsilon) \times \ldots. $$

(a)

Then $U(x,\epsilon)$ is not equal to the open ball $B_{\tilde{\rho}}(x, \epsilon)$ of radius $\epsilon$ centered at $x$; in fact the former properly contains the latter: the point $x^\prime \colon= (x_n + \epsilon- \frac{\epsilon}{n})_{n \in \mathbb{N}}$ is in the former but not in the latter.

Am I right?

(b)

How to show that $U(x, \epsilon)$ is not even open in the uniform topology?

For this purpose, we need to explicitly exhibit a point $y$ in $U(x, \epsilon)$ such that no ball centered at $y$ is contained in $U(x,\epsilon)$. Which point will do, I wonder? I would appreciate if the process could be shown clearly and in detail.

(c) And, I have managed to show that $$B_{\tilde{\rho}}(x, \epsilon) = \bigcup_{0< \delta < \epsilon} U(x, \delta).$$

What is the situation in each of (a), (b), and (c) above (i) if $\epsilon= 1$? and (ii) if $\epsilon > 1$?

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  • $\begingroup$ Note that your point $x'$ from part a) is a reasonable choice for $y$ in part b) $\endgroup$
    – Rolf Hoyer
    Apr 20 '15 at 5:15
  • $\begingroup$ @Saaqib Mahnuud : Try understanding what the open balls are like. $\endgroup$
    – gary
    Apr 20 '15 at 5:58
  • $\begingroup$ @gary, that's just what I'm trying to do but haven't fully so far, I'm afraid. $\endgroup$ Apr 20 '15 at 6:53
  • $\begingroup$ How, @RolfHoyer? To be honest, that's just what I'd expected but haven't quite been able to figure out on my own. $\endgroup$ Apr 20 '15 at 6:54
  • $\begingroup$ If you fix any ball of radius $\epsilon'>0$ around $x'$ you can note that $\epsilon'>\epsilon/n$ for some $n$, which means that you can increase the $n$-th coordinate by enough for the resulting element to not lie in $U(x, \epsilon)$. $\endgroup$
    – Rolf Hoyer
    Apr 20 '15 at 7:11
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Your answer to part (a) is correct. As Rolf Hoyer noted in the comments, you can use the point $x'$ from part (a) to answer part (b): you’ll find that no matter how small a $\delta>0$ you choose, $B_{\tilde\rho}(x',\delta)\setminus U(x,\epsilon)\ne\varnothing$, because, for instance, $x_k'+\frac{\delta}2>x_k+\epsilon$ for all sufficiently large $k$.

If $\epsilon>1$, $B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega$, since $\tilde\rho(x,y)\le 1$ for all $x,y\in\Bbb R^\omega$. $U(x,\epsilon)$, on the other hand, is definitely not all of $\Bbb R^\omega$, so in this case we have $U(x,\epsilon)\subsetneqq B_{\tilde\rho}(x,\epsilon)$. However, $U(x,\epsilon)$ is still not open in the uniform topology: you can use the same basic idea to prove it as in the case $0<\epsilon<1$, taking a point $x'\in U(x,\epsilon)$ such that $x_k'-x_k$ approaches $\epsilon$ as $k$ increases. As far as (c) goes, it’s clear that

$$B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega\supsetneqq\bigcup_{0<\delta<\epsilon}U(x,\delta)$$

when $\epsilon>1$. In fact, by considering a point $x'$ such that $x_k'=x_k+k$ for each $k$, we can see that

$$B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega\supsetneqq\bigcup_{\delta>0}U(x,\delta)\;.$$

Finally, the answers when $\epsilon=1$ are the same as when $0<\epsilon<1$, and essentially the same arguments work.

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  • $\begingroup$ Thank you so much, dear @Brian M. Scott. I've gotten your answer to why $U(x,\epsilon)$ is not open, and I've almost gotten your answer when $\epsilon > 1$, but I've yet to fully convince myself. I'd appreciate if you could replace phrases such as "it is clear that ..." with some more concrete pictures of how something is or is not the case. $\endgroup$ Apr 21 '15 at 18:06
  • $\begingroup$ @Saaqib: When I use it’s clear that, it’s because I really do think that it’s clear, at least after a moment’s thought. Sometimes I’m wrong for a particular reader, by then I’ll be happy to clarify. Here there are two things to check. (1) $B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega$ when $\epsilon>1$: that’s because $\tilde\rho(x,y)\le 1$ for all $y\in\Bbb R^\omega$. (2) $\Bbb R^\omega\supsetneqq\bigcup_{0<\delta<\epsilon}U(x,\delta)$: I actually proved a stronger statement in the next sentence (In fact, ...). The key idea is that if $\lim_{k\to\infty}|x_k-y_k|=\infty$, then ... $\endgroup$ Apr 21 '15 at 18:19
  • $\begingroup$ ... $y$ is not in $U(x,\delta)$ for any $\delta$ at all. $\endgroup$ Apr 21 '15 at 18:19

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