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This is from a review packet:

Let $d:\mathbb{R} \to \mathbb{R}$ be defined as $$d(x,y)=\frac{|x-y|}{1+|x-y|}.$$

i) Show that $(\mathbb{R},d)$ is a bounded metric space.
ii) Show that $A=[a,\infty)$ is a closed and bounded subset of $\mathbb{R}$.
iii) Show that $A=[1,\infty)$ is not compact.

For (i) I think this is a true observation: $d(x,y)\leq1$ for an arbitrary $x,y \in \mathbb{R}$. I'm not sure where to go from there however.

For (ii) and (iii) - I'm assuming those will following quickly from (i).

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  • $\begingroup$ your logic for $(i)$ is correct, The metric $d(x,y)=|x-y|$ is equivalent to $d_1=\frac{d}{1+d}$ $\endgroup$
    – Myshkin
    Oct 25, 2012 at 4:50
  • $\begingroup$ I'm not sure I follow on how to chase definitions into showing that this is indeed a bounded metric space. $\endgroup$
    – emka
    Oct 25, 2012 at 4:53

1 Answer 1

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Since $d(x,y)\leq 1$ for $\forall x \neq y$ we get that $\mathbb R,[a,+\infty)\subset B_1(0)$. Finally $[1,+\infty)$ is not compact since the open cover $\{(0,n), \ n \in \mathbb{N}\}$ does not have a finite subcover of $[1,+\infty)$.

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  • $\begingroup$ I'm not sure I complete follow. I do understand that if $x \neq y$, then $d(x,y)<1$. Though I don't know how to pull that into showing the closed and boundedness and lack of compactness of the sets (respectively). $\endgroup$
    – emka
    Oct 25, 2012 at 6:14
  • $\begingroup$ The definition of a set $A\subset X$ to be bounded is $A\subset B_r(x)$ for some $r>0, \ x \in X$. Since here $\mathbb{R}\subset B_1(0)$, everything is bounded. Now an $A\subset \mathbb{R}$ is open (closed ) with respect to $d$ iff is open (closed ) in the usual sense. This is a consequence of $x_n \to x$ with respect to $d$ iff $|x_n-x|\to 0$. For the compactness we used the definition. $\endgroup$
    – P..
    Oct 25, 2012 at 13:43

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