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Two different metrics $d$ and $\hat d$ in a space $X$ are said to be equivalent iff the topologies generated by them are the same, in other words $U\subseteq X$ is $d$-open iff it is $\hat d$-open. So by definition $(c)$ is correct, am I right?

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2 Answers 2

up vote 9 down vote accepted

No. Even if $d$ and $\hat{d}$ induce the same topology (in which case a set is $d$-closed $\iff$ it is $\hat{d}$-closed), it need not be the case that $\hat{d}$-bounded $\implies$ $d$-bounded, so one cannot conclude from the Heine-Borel theorem ($d$-closed and $d$-bounded $\implies$ $d$-compact for subsets of $\mathbb{R}^n$) that $\hat{d}$-closed and $\hat{d}$-bounded $\implies$ $\hat{d}$-compact for subsets of $\mathbb{R}^n$.

BenjaLim has given the example $\hat{d}(x,y)=\min\{d(x,y),1\}$. The other standard example is $$\hat{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}.$$ One can show that for any metric space $(X,d)$, both of these definitions of $\hat{d}$ induce the same topology as $d$, but that $X$ is bounded under these $\hat{d}$ even if it was not under $d$.

Remember that boundedness is strictly a property that comes from a metric, i.e. boundedness is not a topological property.

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example please.. –  Une Femme Douce Jul 22 '12 at 23:36
    
Just an idea: what if $\hat{d}$ would be such that every bounded set would also be totally bounded respect to this metric (as in the Euclidean case). Would this be sufficient for these statements to hold? And would it also be a necessary condition? –  Thomas E. Jul 23 '12 at 4:15
    
@ZevChonoles +1 For showing another example to Patience :D –  user38268 Jul 25 '12 at 22:44

(c) is false, as an example take $\hat{d}(x,y) = \min\{d(x,y),1\}$. Then you can check that $\hat{d}$ and $d$ generate the same topology. Now $\Bbb{R}$ is closed and bounded with respect to the $\hat{d}$ metric, but clearly it cannot be compact: Any open cover of $\Bbb{R}$ in the $\hat{d}$ metric is also an open cover for $\Bbb{R}$ in the $d$ metric which has no finite subcover.

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