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I asked a question last night about proving that a discrete metric space is both open and closed. Once or twice it was mentioned that a set that contains only one element is open. I'd like to know:

a) is that always true?

b) why?

An explanation that is both a proof and a simple breakdown of it would be most helpful.

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    $\begingroup$ The proof is little more than "What is the definition of 'open'? What is the definition of 'discrete'?" $\endgroup$ – Hurkyl Aug 27 '13 at 18:06
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    $\begingroup$ It's true in discrete spaces (every subset of a discrete space is open and closed), not in general. However, in general, a singleton set is open iff the point it contains is an isolated point. $\endgroup$ – Daniel Fischer Aug 27 '13 at 18:07
  • $\begingroup$ he may want to know why the discrete metric space has a discrete topology. $\endgroup$ – magguu Aug 27 '13 at 18:15
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    $\begingroup$ @magguu *she and no. $\endgroup$ – Siyanda Aug 27 '13 at 18:17
  • $\begingroup$ Any topological space is both open and closed. One of the definitions of a topology is that the empty set and the entire set are both open sets, and therefore both closed. You may need to change the first sentence of your question. $\endgroup$ – Stefan Smith Aug 28 '13 at 1:08
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In general, it is not the case that single-element sets are open. For example, in the usual topology on $\Bbb R$, the one induced by the usual metric, single-element sets are not open; open sets are unions of open intervals, and every open set (except $\varnothing$) is infinite.


Say we have a metric space with the so-called "discrete" metric $d$. Recall that this means that $$d(x,y) = \begin{cases} 0, \text{if $x = y$}\\ 1, \text{if $x\ne y$}\end{cases}$$

A metric space has a natural topology "induced by" its metric. The "metric topology" induced by the metric $d$ is the one that has as its basis all "balls" $N_{d,\epsilon}(x)$ where $$N_{d,\epsilon}(x) = \{ p \mid d(x,p) < \epsilon\}.$$

"Basis" here means that a set is open if and only if it is a union of some of these balls.

The topology induced on $\Bbb R^n$ by its usual metric is exactly the usual topology for $\Bbb R^n$. But for an unusual metric such as the discrete metric, the situation is different.

Observe that in the discrete metric $d$, $N_{d,1/2}(x) = \{x\}$, because $x$ itself is the only point $p$ such that $d(x,p) < \frac12$. So each $\{x\}$ is a basis element and is therefore open in the metric topology induced by $d$.

Since any union of open sets is open, and any set at all is a union of sets of the form $\{x\}$, we conclude that any set at all is open in the topology induced by the discrete metric $d$.

This is why this metric $d$ is called the "discrete" metric: the topology it induces is the discrete topology.

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I am not sure whether you mean a) and b) in general or just in a discrete metric space so let me answer for both cases

A) for the discrete metric space it is always true by definition. for a general topology it is not

B) consider the ball $B(x,1/2)$. it is a basic open set and contains only one point $x$. if the space is not discrete metric then Real numbers give you an example of why a singleton is not open.

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