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Are these the same concepts?

I am particularly interested in space-time with a discrete structure and the like. I heard that there is not a standard way for quantizing space-time yet. At any rate, I would like also to know in a quantized space-time, coordinates merely take discrete values? Or they take discrete values only in some special quantum mechanical conditions?

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    $\begingroup$ The mathematical definition is that a discrete space has the discrete topology: every point is an open set. The integers have this property, the rationals do not. $\endgroup$ – Javier Oct 2 '16 at 17:09
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Oct 2 '16 at 18:00
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Defintion of a discrete set: click

Definition of a countable set: click

Finite sets are both, countable and discrete.

The set $$ S := \{ \frac{1}{n} ~ | ~ n \in \mathbb{N} \} $$

is countable and discrete.

The set

$$ S' = \overline{S} = S \cup \{0\} $$

is countable but not discrete.

Proof: let f: $S' \rightarrow \mathbb(N) $ with $f(0) = 1$ and $f(1/n) = n+1$ then f is injective. Thus $S'$ is countable. However, $S'$ is closed and has an accumulation point $0 \in S'$. Thus it cannot be discrete (see definition).

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    $\begingroup$ Thanks for answer. The set S is a subset of $\mathbb Q$. Is there any general formula for presenting countable and discrete subsets of $\mathbb Q$? $\endgroup$ – user.3710634 Oct 2 '16 at 19:37
  • $\begingroup$ @J.Pak: The rational numbers $\mathbb{Q}$ with it's usuall topology relative to $\mathbb{R}$ are not discrete. Subsets can be but I suppose there is no general rule to that. You just have to proof or disproof it for each set. Also, the fact that S or S' being discrete (or not) has nothing to do with them being a subset of $\mathbb{Q}$. Take for instance $S+\pi$, then no element is rational, yet they are topological equivalent. Every subset of $\mathbb{Q}$ is countable though (because $\mathbb{Q}$ is). $\endgroup$ – image Oct 2 '16 at 23:01
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I would understand discrete such that there is a minimum distance between the nearest element. A cubic lattice is an example of this.

In mathematics, the expression countable means that there exists a mapping from the natural numbers to the set you describe. This means that you can sequentially number every point. In a cubic lattice, this is certainly possible with an appropriate scheme. So a cubic lattice (as in an ordinary solid crystal) would serve as a countable and discrete space.

The rational numbers $\mathbb Q$ are also countable. Yet I would not call it discrete because you can always find (countable) infinitely many numbers between any two numbers. If you would make a spacetime which was $\mathbb Q^{1,3}$ instead of the usual $\mathbb R^{1,3}$ it would be countable but I wouldn't call it discrete.

A discrete spacetime is used in K. Wilson's lattice field theory where it serves as a regulator. The coordinates there do take discrete values, you just number the lattice sites with numbers from $\mathbb N^4$.

In quantum field theory, spacetime is not quantized. It is also uncountably infinite.

About the “no standard way”: Discretizing the space with a lattice is a computation aid. It has no physical interpretation. It is just darn useful when doing simulations on a computer!

There are thoughts that spacetime might be discrete in a physical way. One of this is quantum loop gravity which discretizes space at the order of $10^{-34} \, \mathrm m$. This is so small that nobody has any idea what could happen there.

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    $\begingroup$ A discrete set need not have a minimum distance, see my answer. Also countable does not mean that there is a map from $\mathbb{N}$ to the set. It's the other way round + that map has to be injective. $\endgroup$ – image Oct 2 '16 at 17:47
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"Countable" and "discrete" are separate concepts. The Cantor set, for example, is discrete, but not countable.

As for discretizing space-time, the only theory I'm aware of that takes this approach is known as loop quantum gravity. String theory, if my passing understanding is correct, does not discretize space-time in this way.

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    $\begingroup$ The cantor set is not discrete. The cantor set is closed. A closed discrete set cannot have any accumulation points. However, $1/4 = 0.0202020202... $ (ternary) is an accumulation point. $\endgroup$ – image Oct 2 '16 at 17:28
  • $\begingroup$ Am I confusing discrete with disconnected, then? $\endgroup$ – Sean E. Lake Oct 2 '16 at 17:29
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    $\begingroup$ Probably. The cantor set is disconnected $\endgroup$ – image Oct 2 '16 at 17:33

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