How do i geometrically visualise discrete metric space?

Question

This is a informal question i guess. Some metrics is easy to understand intuitively and draw some understandable diagrams denoting convergence or any metric property.How can i draw a discrete metric ?

Answer 1

I like to think of it this way.

Metrics always tell us something about "distance" between two points in a space. While distance can have varying definitions, there is some measure of closeness that a metric shows us.

Now we can see that the discrete really just tells us whether $x=y$ or not when we have $d(x,y)$ as it outputs $0$ for $x=y$ and $1$ for $x\neq y$. So we can intuitively understand the discrete metric as a machine of sorts that tells us whether two points are the same or not. I cannot think of a geometric interpretation, the "Same-or-not Machine" is the best way I find to think about the discrete metric.

Answer 2

A metric space is discrete iff there is no point a and sequence of points {$a_j$}, none of which are a, for which the sequence d(a,$a_j$) converges to 0.

A space is discrete iff there is no point a and net n,
no points of which are a, for which n converges to a.

If the space is first countable, then sequences instead
of nets can be used.

Answer 3

From the topological point of view the mutual distances between points don't all have to be $1$. As long as they are bounded away from $0$ the topology is the same. So just sprinkle the points around in the plane for a picture. (You will encounter a problem if there are more than uncountably many points, but that shouldn't get in the way of your intuition.)

Answer 4

The term discrete metric space is actually ambiguous. More precisely, it is ambiguous geometrically but not topologically. Here are three possible definitions of "the metric space $(X,d)$ is discrete":

1) $d$ is the discrete metric $d_0$ on the set $X$: i.e., $d_0(x,y) = \begin{cases} 0 & x = y \\ 1 & x \neq y \end{cases}$.

2) $d$ is "uniformly discrete": the metric $d$ is uniformly equivalent to the discrete metric $d_0$ on $X$: in abstract terms, this means that the identity function $1_X: X \rightarrow X$ is uniformly continuous as a function from $(X,d_0)$ to $(X,d)$ and also as a function from $(X,d)$ to $(X,d_0)$. This immediately unpacks as follows: for all $\epsilon > 0$ there are $\delta_1, \delta_2 > 0$ such that for all points $x,y \in X$, we have

$$d(x,y) \leq \delta_1 \implies d_0(x,y) \leq \epsilon \text{ and } d_0(x,y) \leq \delta_2 \implies d(x,y) \leq \epsilon.$$

3) $d$ is "topologically discrete": the topological spaces $(X,d)$ and $(X,d_0)$ are homeomorphic: i.e., the identity function $1_X$ is continuous as a function from $(X,d)$ to $(X,d_0)$ and also as a function from $(X,d_0)$ to $(X,d)$.

Then we have 1) $\implies$ 2) $\implies$ 3).

In terms of how to visualize these various notions of discreteness:

3) A metric space $(X,d)$ is topologically discrete if and only if for every point $x$ of $X$, there is some $\delta(x) > 0$ such that the open ball $B_x(\delta(x))$ centered at $x$ with radius $\delta(x)$ is equal to $\{x\}$. If this occurs, then indeed $\mathcal{C} = \{B_x(\delta(x)\}_{x \in X}$ expresses $X$ as a disjoint union of open balls, each consisting of a single point. Thus one can view a topologically discrete metric space as being a bunch of "disjoint bubbles": every point is enclosed in a bubble of its very own, and there is no way for distinct points to interact with each other in any topological way. (As an aside, $\mathcal{C}$ is an open covering of $X$ without any proper subcovering. A topological space has such a covering if and only if it is discrete.)

2) A metric space $(X,d)$ is uniformly discrete if and only if we can choose $\delta(x)$ for all $x \in X$ such that $\delta(x) \geq \delta$ for some fixed positive $\delta$. (Taking $\epsilon = 1$ in the definition of uniformly discrete shows that this condition is necessary; it is easy to see that it is sufficient.) Thus not only does each point have a bubble of its own, but (intuitively speaking) there is enough room between the points that the bubbles do not get too small in size.

1) The discrete metric itself is a case of 2) in which the largest possible bubble one can place around each point has radius $1$. Admittedly it is sort of difficult to picture this truly geometrically: the largest number of points in the Euclidean space $\mathbb{R}^N$ that are mutually equidistant from one another is $N+1$, so to think of an infinite discrete metric in Euclidean terms one needs an infinite dimensional Euclidean space. (And indeed a suitable Hilbert space would suffice -- good for you if this helps with your geometric intuition! Not so much for me...)

An example of a metric space that is topologically discrete but not uniformly discrete is $S = \{\frac{1}{n} \mid n \text{ is a positive integer}\}$ endowed with the restriction of the usual Euclidean metric: it is easy to see that the radius of the bubble around $\frac{1}{n}$ must approach $0$ as $n$ appraoches $\infty$.

A uniformly discrete metric must share the same "uniform properties" as the discrete metric. (This can be formalized in terms of uniform spaces, but this particular gadget lies outside the vocabulary of most contemporary mathematicians and students.) For example, every uniformly discrete metric space is complete: all Cauchy sequences converge. Indeed, every Cauchy sequence is eventually constant, since in a uniformly discrete space, there is a $\delta > 0$ such that no open ball of radius at most $\delta$ contains more than one point! However the topologically discrete metric space $S$ above is not complete, being defined as a non-closed subset of $\mathbb{R}$. Since $\mathbb{R}$ is complete, its completion is $S \cup \{0\}$. (Maybe it is interesting to ask what class of metric spaces one gets by completing topologically discrete metric spaces? I haven't given it much thought.)

Finally, although topologically discrete spaces (of the same cardinality) all look the same topologically, while all uniformly discrete metric spaces are all uniformly equivalent, they can have very different metric properties. For instance, a uniformly discrete metric space on the underlying set $\mathbb{Z}$ can be either bounded (the discrete metric) or unbounded (the Euclidean metric). The finer relation of Lipschitz equivalence preserves boundedness. To any finitely presented group $G$ one can attach a countable, uniformly discrete metric on $G$ and the study of such uniformly discrete metrics up to quasi-isometry (which is not so different from Lipschitz equivalence) is an entire branch of mathematics, geometric group theory.

As another example, a lattice in $\mathbb{R}^N$ is the set of $\mathbb{Z}$-linear combinations of an $\mathbb{R}$-basis for $\mathbb{R}^N$: otherwise put, every lattice is the image of the standard lattice $\mathbb{Z}^N$ under an invertible linear transformation of $\mathbb{R}^N$. All lattices in $\mathbb{R}^N$ are Lipschitz equivalent to each other (the strongest notion of equivalence discussed here) and they are all uniformly discrete. But they certainly have very different and interesting geometric properties, and the study of this is a large part of the branch of mathematics known as, um, discrete geometry.

Answer 5

Using a line segment of length one for 1D, equilateral triangle with unit length sides for 2D, a tetrahedron with unit length sides for 3D.

I was wondering the same to explain it somehow to the students possibly that every open sphere in a discrete space contains only its center. This can be explained by visualizing the metric space in 3D containing points as nodes/vertices of a tetrahedron all at distance one from each other (hence satisfying the discrete space condition easily). In this way, I could explain that any of the other nodes cannot be reached by the sphere with radius less than the length of any of the edges joining those points (edges are there just for explaination of relative distance reached).

You may use such a thing to visualize the discrete space.