Exploring Open, Closed, Bounded, and Compact Sets in R with a Unique Metric

[Silviu]

Homework Statement

Consider the set of real number with the following metric: $\frac{|x-y|}{1+|x-y|}$. Which subsets of R with this metric are open, closed, bounded or compact?

Homework Equations

The Attempt at a Solution

First I calculated the neighborhood in this metric. If the radius of the neighborhood is bigger than 1, it contains the whole R (this implies that all sets are bounded). If a neighborhood has a radius smaller than 1, that it is an interval of the form $(a-\frac{r}{1-r},a-\frac{r}{1+r})$. Now looking at open sets, a set is open if for every element there is a neighborhood of that element completely in the set and this allows us to take r as small as we want, so even smaller than 1. So if open sets are made of open intervals (open intervals in R with the normal metric), and the union of open intervals is an open interval, it means that open sets are open intervals. For closed sets and compact, we can use the same reasoning as in R with the normal metric as we can take r < 1 and make the substitution $r \to \frac{r}{1-r}$. So closed and compact sets are the same as in R. Is my reasoning good enough, like is it rigorous, or there are some spaces that need to be filled? Thank you!

 

[Bashyboy]

It could use some polishing, but I think you are heading in the right direction. I suppose the rigor of your proof hinges upon these calculations you performed but didn't include. Let $g(x,y) = \frac{|x-y|}{1+|x-y|}$ and let $d(x,y) = |x-y|$, which is the standard metric. For boundedness, you have to be careful. A subset nonempty subset $S$ of real numbers is bounded provided there exists some $S > 0$ such that $g(x,y) \le M$ for all $x,y \in S$. So I don't think you actually proved that all sets are bounded, although you are right that all sets are bounded (in fact, $g(x,y)$ is typically called the standard bounded metric).

Regarding the closed and compact sets, there isn't really anymore work to do once you have shown that open sets generated by the (bounded) metric $g$ are the same as those in the generated by the standard (Euclidean) metric $d$. To do this, given

$$B_g(x,\epsilon) := \{y \in \Bbb{R} \mid g(x,y) < \epsilon \} = \{ y \in \Bbb{R} \mid \frac{|x-y|}{1 + |x-y|} < \epsilon \}$$ (i.e., the $\epsilon$ ball with respect to the metric $g$), you need to find a $\delta > 0$ such that

$$B_{d}(x,\delta) \subseteq B_g(x,\epsilon),$$

where $B_d(x,\delta) = \{ y \in \Bbb{R} \mid d(x,y) < \delta \} = \{y \in \Bbb{R} \mid |x-y| < \delta \}$. Then you have to do the samething with the roles of $g$ and $d$ reversed (make sure you draw lots of pictures--they will help you find the $\delta > 0$ and $\epsilon > 0$ you need).

Perhaps you ought to include a precise formulation of your definitions of open, closed, and compact sets.