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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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    I'm not sure I follow on how to chase definitions into showing that this is indeed a $b$ounded metric space.2012-10-25

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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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    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.2012-10-25