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Let $\overline{\rho}$ be the uniform metric on $\Bbb{R}^\omega$. Given $x=(x_1, x_2,….)\in \Bbb{R}^\omega$ and given $0\lt \epsilon \lt 1$, let $U(x,\epsilon)=(x_1 -\epsilon ,x_2 +\epsilon )\times \dotsb \times (x_n -\epsilon ,x_n +\epsilon) \times \dotsb$. (a) show that $U(x,\epsilon)$ is not equal to the $\epsilon$ ball $B_{\overline{\rho}}(x,\epsilon)$. (b) show that $U(x,\epsilon)$ is not even open in the uniform topology. (c) show that $B_{\overline{\rho}}(x,\epsilon)=\bigcup_{\delta \lt \epsilon}U(x,\delta)$.

My attempt: (a) let $y\in B_{\overline{\rho}}(x,\epsilon)$. Then $\overline{\rho}(x,y)\lt \epsilon \lt 1$, last inequality follows from hypothesis of exercise. So $\overline{d}(x_i,y_i)= d(x_i,y_i)\lt \epsilon ,\forall i\in \Bbb{N}$. So $y_i \in (x_i -\epsilon ,x_i +\epsilon), \forall i\in \Bbb{N}$. Thus $y=(y_n)\in \prod_{i\in \Bbb{N}}(x_i -\epsilon, x_i +\epsilon)=U(x,\epsilon)$. Hence $B_{\overline{\rho}}(x,\epsilon) \subseteq U(x, \epsilon)$, if $0\lt \epsilon \lt 1$.

Conversely, let $y\in U(x,\epsilon)$. $y_i \in (x_i -\epsilon ,x_i +\epsilon), \forall i\in \Bbb{N}$. So $d(x_i,y_i) \lt \epsilon \lt 1$. By definition, $\overline{d}(x_i,y_i)=d(x_i,y_i) \lt \epsilon ,\forall i\in \Bbb{N}$. Thus $\overline{\rho} (x,y)\leq \epsilon$. There is a possibility that $\overline{\rho}(x,y)=\epsilon$, in that case $y\notin B_{\overline{\rho}}(x,\epsilon)$. Hence $U(x, \epsilon)\nsubseteq B_{\overline{\rho}}(x,\epsilon)$. Is this proof correct?

(b) I don’t see a general way to prove this part.

(c) we first prove a general result. If $0\lt r \lt r’$, then $U(x,r)\subseteq B_{\overline{\rho}}(x,r’)$. Proof: let $y\in U(x,r)$. Then $d(x_i,y_i)\lt r, \forall i\in \Bbb{N}$. So $\overline{d}(x_i,y_i)\leq d(x_i,y_i)\lt r ,\forall i$. Thus $\overline{\rho}(x,y)\leq r\lt r’$. Hence $y\in B_{\overline{\rho}}(x,r’)$. Proof is similar to theorem 20.4 proof, $\mathcal{T}_{\overline{\rho}}\subseteq \mathcal{T}_b$ part.

Let $y\in \bigcup_{\delta \lt \epsilon} U(x,\delta)$. Then $y\in U(x,\delta’)$, for some $\delta’ \lt \epsilon$. Thus $y\in U(x,\delta’)\subseteq B_{\overline{\rho}}(x,\epsilon)$, by above claim.

Conversely, let $y\in B_{\overline{\rho}}(x,\epsilon)$. Then $\overline{\rho}(x,y) \lt \epsilon \lt 1$. $\exists \delta \in \Bbb{R}$ such that $\overline{\rho}(x,y)\lt \delta \lt \epsilon$. Thus $\overline{d}(x_i,y_i)=d(x_i,y_i) \lt \delta ,\forall i \in \Bbb{N}$. Hence $y \in U(x,\delta)\subseteq \bigcup_{\delta \lt \epsilon} U(x, \delta)$. Is this proof correct?

user264745
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  • How are you showing in part (a) that $U(x,\varepsilon) \neq B_{\bar{\rho}}(x,\varepsilon)$? For that you need to define a point $p$ that lies in the first set, but not in the second or vice versa. Where are you defining this point? Your proof of one inclusion is correct but to show non-inclusion you need an explicit counterexample. – Henno Brandsma Feb 05 '22 at 08:01
  • See my answer for inspiration. – Henno Brandsma Feb 05 '22 at 08:06
  • See also this? Search first – Henno Brandsma Feb 05 '22 at 08:07
  • @HennoBrandsma I think there are lots of ways to construct an explicit counterexample. I’m trying to prove part(a) without constructing any counterexample. I want to show $U(x,\epsilon) \nsubseteq B_{\overline{\rho}}(x,\epsilon)$, so $U(x,\epsilon) \neq B_{\overline{\rho}}(x,\epsilon )$. In proof of that inclusion I concluded, in general $\overline{\rho}(x,y)$ might be equal to $\epsilon$. So $y\notin B_{\overline{\rho}}(x,\epsilon)$. In all the counterexample construction the point one choose indeed satisfy $\overline{\rho}(x,y)=\epsilon$ condition. – user264745 Feb 05 '22 at 08:52
  • @HennoBrandsma https://math.stackexchange.com/a/2770137/861687 this is my fourth way to construct a counterexample. – user264745 Feb 05 '22 at 08:55
  • You cannot show (a) without a counterexample. That’s my whole point. You only sketch what a counterexample should look like, but you have to give an explicit example. – Henno Brandsma Feb 05 '22 at 09:39
  • @HennoBrandsma Is it(counterexample) also true for part(b) of the problem? – user264745 Feb 05 '22 at 11:11
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    Yes for b you have to give a point that’s not interior so an example is needed. The same point as a will probably work. – Henno Brandsma Feb 05 '22 at 12:04
  • @HennoBrandsma Thank you for clarifying that thing. – user264745 Feb 05 '22 at 12:39

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