Let $t \in \mathbb{Z}^+ := \mathbb{N} \cup \{0\}$ and let $\beta > 1$ be a fixed parameter. For $a, b \in \mathbb{R}$, define $\rho(a, b) = \min\{|a-b|, 1\}$. Then for any $x, y \in \mathbb{R}^{\infty}$ (here $\mathbb{R}^{\infty}$ represents the infinite Cartesian product of $\mathbb{R}$ with itself, i.e., $\mathbb{R} \times \mathbb{R} \times \cdots$), define $$d(x, y) = \sum_{t \in \mathbb{Z}^+} \beta^{-t}\rho(x_t, y_t) $$ as the "distance" between $x = (x_0, x_1, \cdots)$ and $y = (y_0, y_1, \cdots)$. Show that $(\mathbb{R}^{\infty}, d)$ is a bounded metric space.

First, I have already proved that $d$ is a valid metric on $\mathbb{R}^{\infty}$. I know the definition of a bounded metric space is

Let $(X, d)$ be a metric space. A subset $S \subseteq X$ is bounded if $\exists x \in X$, $\exists \varepsilon>0$ such that $A \subseteq B_{\varepsilon}(x)$.

But I am unsure how to apply that definition here to prove that $(\mathbb{R}^{\infty}, d)$ is bounded. Any help would be appreciated!

EDIT: In a related problem, I am asked to prove whether $[0,1]^{\infty}$ (the infinite Cartesian product of $[0,1]$ with itself) is an open or closed subset of $\mathbb{R}^{\infty}$. I know the definition that for a metric space $(X, d)$, a set $A \subseteq X$ is open if for all $x \in A$, there exists a $\varepsilon >0$ such that $B_{\varepsilon}(x) \subseteq A$. A set $C \subseteq X$ is closed if and only if $X \setminus C$ is open. How can I apply that here?

  • $\begingroup$ Hint on $[0, 1]^\infty$ question: suppose $x \notin [0, 1]^\infty$. Then there exists $t \in \mathbb{Z}^+$ such that $x_t < 0$ or $x_t > 1$. In the second case, what can you say about the ball around $x$ with radius $\beta^{-t} (x_t - 1)$? $\endgroup$ – Daniel Schepler Jul 26 '17 at 23:07
  • $\begingroup$ (Or, for a slightly less direct approach: show that for each $t$, the projection map $\pi_t : \mathbb{R}^\infty \to \mathbb{R}$ is continuous with respect to $\rho$ on the domain and the normal metric on the codomain. Then you can express $[0, 1]^\infty$ as $\bigcap_{t=0}^\infty \pi_t^{-1}([0, 1])$.) $\endgroup$ – Daniel Schepler Jul 26 '17 at 23:09
  • $\begingroup$ Hmm, with your first point, I can see what you're trying to do, so you are working with the set $\mathbb{R}^{\infty} \setminus [0,1]^{\infty}$ and then trying to see whether it is closed/open to deduce whether $[0,1]^{\infty}$ is open or closed. But I am not sure how to continue your argument, can you provide the finish? $\endgroup$ – elbarto Jul 27 '17 at 1:17
  • $\begingroup$ OK, the point is: if $d(x', x) < \beta^{-t} \rho(x_t, 1)$, then $d(x', x) \ge \beta^{-t} \rho(x_t, x_t')$ so $\rho(x_t, x_t') < \rho(x_t, 1)$ and that implies $x_t' > 1$ so $x' \notin [0, 1]^\infty$. So there's a small correction that the radius should be $\beta^{-t} \rho(x_t, 1)$. (You now just have to prove the lemma: if $x > 1$ and $\rho(x, y) < \rho(x, 1)$ then $y > 1$, which should be straightforward.) $\endgroup$ – Daniel Schepler Jul 27 '17 at 16:46
  • $\begingroup$ Ah I see, and what about for $x_t<0$? $\endgroup$ – SwiftMo Jul 27 '17 at 22:31

$d(x,y)\leq \sum_{t\in\mathbb{Z}^+}\beta^{-t}$. Write $c=\sum_{t\in\mathbb{Z}^+}\beta^{-t}$. Take any $x_0\in\mathbb{R}^{\infty}$, $d(x_0,y)\leq c$ for every $y$.

  • $\begingroup$ Thanks, I edited my OP with a related problem, that is whether $[0,1]^{\infty}$ is open or closed, could you have a look? I know the definitions but not sure on how to apply it here. $\endgroup$ – elbarto Jul 26 '17 at 22:58

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