Let $\mathbb{R}^{\omega}$ denote the product of countably-many copies of $\mathbb{R}$. Let $\bar{d}$ be the following metric on $\mathbb{R}$: $$\bar{d}(x,y)=\inf\{|x-y|,1\}$$ The uniform metric on $\rho$ on $\mathbb{R}^{\omega}$ is given as $$\rho((x_{n}),(y_{n}))=\sup\{\bar{d}(x_{n},y_{n}):n \in \mathbb{N}\}$$ For a point $x=(x_{n}) \in \mathbb{R}^{\omega} $ and $\epsilon > 0$ given, define $$U(x,\epsilon)=(x_1 - \epsilon,x_1 + \epsilon)\times...\times(x_n - \epsilon,x_n + \epsilon)\times...$$ and the open $\epsilon$-disk $B_{\epsilon}(x)$ ($=$ open $\epsilon$ ball) in the uniform metric.

My questions are as follows:

  1. Show that $U(x,\epsilon)$ is not even open in the uniform topology.

My idea was to show that there is a point in a ball that would be contained in the set $U(x,\epsilon)$, that at the same time, "sticks out" of $U(x,\epsilon)$. The way to do this, I would guess, would be to consider some sort of sequence of coordinates that converges to something that is clearly not in $U(x,\epsilon)$, but does not necessarily violate the restriction on distance in regards to the uniform metric. I know that the box topology that $U(x,\epsilon)$ pertains to is as fine as it gets, so given any $\epsilon$ I should be able to carry out this idea, but I have not been able to come up with a point or, equivalently, a sequence, that actually works.

  1. Show that $B_{\epsilon}(x)=\bigcup_{\delta<\epsilon} U(x,\delta)$

I do not have any work for this and am unsure where to start.


Your intuition is right for the first question. Let $y_n = x_n + (1 - 2^{-n})\epsilon$; then clearly $y = (y_n) \in U(x,\epsilon)$. But no ball around $y$ in the uniform metric is contained in $U(x,\epsilon)$: if the radius of such a ball is $\delta$ and $\delta > 2^{-n}\epsilon$ for some $n$, then $B_\delta(y)$ contains, for example, $z = (z_k)$ where $$z_k = \begin{cases} y_k & k \ne n \\ y_n + 2^{-n}\epsilon & k = n, \end{cases}$$ which is not in $U(x,\epsilon)$.

The second question is asking you to check two things. The first is that every $U(x,\delta)$ for $\delta < \epsilon$ is contained in $B_\epsilon(x)$. That is, every point $y$ in $U(x,\delta)$ has $\rho(x,y) < \epsilon$. But in fact, $\rho(x,y) \le \delta$, since $|y_n - x_n| < \delta$ for each $n$. (Make sure you understand why this is $\le \delta$ and not $< \delta$ -- this is the same principle behind my answer to the first question.) The second thing to check is that every $y$ with $\rho(x,y) < \epsilon$ is contained in some $U(x,\delta)$. If $\rho(x,y) = \epsilon' < \delta < \epsilon$, then you can easily check that $y$ is contained in $U(x,\delta)$.

Here's some 'philosophy' that may be helpful. Membership in these box-topology-style open sets like $U(x,\epsilon)$ is determined by each coordinate individually, but membership in uniform open sets is determined by the behavior of the entire sequence $(y_n)$, since you're taking a supremum.

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  • $\begingroup$ I've been looking up for the answer of the question which is the same as the one asked here and arrived here. Shouldn't one choose n such that 2^(-n) is smaller than the minimum of the delta and 1 since the standard bounded metric is used in the definition of the uniform metric? Just out of curiosity. Thanks. $\endgroup$ – Mathcho Jul 17 '15 at 17:20
  • $\begingroup$ @Paul VanKoughnett Why are you allowed to choose $\delta > 2^{-n} \epsilon$ for some $n$? Isn't $\delta$ suppose to be arbitrary? $\endgroup$ – user193319 Mar 2 '17 at 19:18

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