We can read here that every discrete metric space (where the topology is the same as the discrete topology, i.e. where all the singletons are open) is complete, but an example bothers me because I don't see where I am wrong :
The space $(\{\frac{1}{n}\,;\,n\in\mathbb{N^*}\},d)$ where $d(x,y)=|y-x|$ is discrete because you have for all $n\in\mathbb{N}^*$ that
$$B\Big(\frac{1}{n},\varepsilon_n\Big)\cap S=\Big\{\frac{1}{n}\Big\},\text{ where }\varepsilon_n=\frac{1}{2n(n+1)}\text{ for example.}$$
But this space is not complete, as the sequence $(\frac{1}{n})_{n\geq 1}$ is Cauchy but doesn't converge in $S.$ My questions are the following :
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Did I tell something wrong in my example ?
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Is the property only true for uniformly discrete metric spaces, where we don't allow $\varepsilon_n$ to become as little as it wants as in this example ?
Thank you for your help !
Best Answer
We have to be careful and distinguish between
a) a metric space that happens to induce the discrete topology (i.e., where each singleton set is open, such as in your example) and
b) a discrete metric space, that is a set endowed with the discrete metric $$d(x,y)=\begin{cases}0&x=y\\1&x\ne y.\end{cases}$$
Of course, the topology induced by the discrete metric is also the discrete topology, but two metrics defining the same topology need not be equivalent. Your very example is excellent to remind us that some properties of metric spaces (such as completeness) are really properties of the metric space, not of the induced topology: The set $\{\,\frac1n:n\in\Bbb N\,\}$ endowed with the metric inherited from $\Bbb R$ is discrete, but is not a discrete metric space; and it is not complete with respect to this metric. Endowed with the discrete metric, however, it becomes complete (without changing its topology): In a discrete metric space every Cauchy sequence is eventually constant, hence convergent.
We might say that discrete metric space should be understood as being "linguistiucally parenthesized" as (discrete metric) space and not as discrete (metric space). I am not sure if it would we ok to write discrete, metric space for a metric space with discrete topology; and even if so, I'd rather be more explicit about the distinction than hiding it in the presence or absence of a comma. Thus I'd prefer to say "metric space with discrete topology" and "space endowed with the discrete metric" if the distinction is needed. The underlying problem may be that the associative law does not hold for natural languages. The standard example of the German language for this is Mädchenhandelsschule which is (of course) intended to mean Mädchen+(handels+schule) ("business school for girls) and not (Mädchen+handels)+schule ("school where you learn girl trafficking")