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In my book, it says that any open ball $B(a,r)$ over the real numbers is equal to the open interval $(a-r,a+r)$. I wonder how I can prove that this is true, only using the metric axioms.

If the metric equals $d: \mathbb{R} \times \mathbb{R}: a,b \mapsto |b-a|$, then this is obviously true.

I know that a metric represents the distance, but does it necessarily have to equal the actual distance function $d$ I mentioned above. The reason why I ask this is because on $R^n$ you have several metrics, e.g. the Euclidian metric, the maximum metric, ...

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up vote 8 down vote accepted

The statement is predicated on the assumption that $B(a,r)$ is defined in terms of the Euclidean metric. The function $$d:\mathbb{R}\times\mathbb{R}\to\mathbb{R}:(x,y)\mapsto 2\vert x-y\vert$$ is a perfectly good metric on $\mathbb{R}$ that generates the usual topology, but $$\begin{align*} B_d(a,r) &= \{x\in\mathbb{R}:d(a,x)<r\}\\ &= \{x\in\mathbb{R}:2\vert a-x\vert<r\}\\ &= \left(a-\frac{r}2,a+\frac{r}2\right), \end{align*}$$

not $(a-r,a+r)$.

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Ok, thanks you cleared that up! I'll accept your answer as soon as the system lets me. –  sxd Oct 2 '11 at 1:23

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