2

Find three non equivalent metrics on infinite set.

For case infinite countable set we can use this answer for rational numbers.

Also for example in $ \mathbb {R}$ we can theree non equivalent metric : discrete metric, euclidean metric and $p-adic $ metric.

But two infinite uncountable set don't haven same cardinality and we can't use idea for case countable set.

How can I find three non equivalent metrics on infinite uncountable set ?

amir bahadory
  • 4,756
  • 1
    How do you define the $p$-adic metric on the reals? I guess $\mathbb Q_p$ has the same cardinality as the reals. – Thomas Andrews Feb 15 '24 at 17:29
  • ...you found three inequivalent metrics on the infinite uncountable set $\mathbb{R}$, so what is your question? Also what do you mean "two infinite uncountable sets don't have the same cardinality"? $\mathbb{R}$ and $\mathbb{C}$ are not the same set, but their cardinality is the same. – Ben Steffan Feb 15 '24 at 17:39

2 Answers2

4

You didn't specify what kind of equivalence you are talking about, but I will assume you mean topological equivalence. Regardless, if two metrics are not topologically equivalent, then they cannot be equivalent in any other reasonable sense.

Instead of defining three such metics on a set $X$ we can define three metrics on some three sets, each of cardinality $|X|$. If we manage to make those spaces non-homeomorphic then we are done.

The topological invariant we are going to look at is: the number of non-isolated points. Since every homeomorphism has to map non-isolated points to non-isolated points, then if two spaces $X$, $Y$ have different number/cardinality of non-isolated points then they cannot be homeomorphic.

So how do we do that on an arbitrary infinite set $X$? First we generate a countable metric space $Y_n$ with $n$ non-isolated points. To do that we can first define $A_n=\{(n,1/k)\ |\ k\in \mathbb{N}\}\cup\{(n,0)\}$ and we put $Y_n=\bigcup_{i=1}^n A_i$. Then $Y_n$, with Euclidean metric, has exactly $n$ non-isolated points, namely: $(1,0),(2,0),\ldots,(n,0)$.

Then we take the disjoint union $X\sqcup Y_n$, and extend the Euclidean metric from $Y_n$ to whole $X\sqcup Y_n$ by putting $d(x,y)=1$ whenever $x\neq y$ and one of them is in $X$. This adds only isolated points to $Y_n$, and therefore $X\sqcup Y_n$ is a metric space of cardinality $|X|$ with exactly $n$ non-isolated points. Meaning $X\sqcup Y_n$ is not homeomorphic to $X\sqcup Y_m$ for $n\neq m$, yet they all are of cardinality $|X|$.

So here you go. I even gave you countably many such examples.

freakish
  • 47,446
2

@freakish answer is alright but I find this simpler:

Given a cardinal $\kappa$, take two discrete spaces $X, Y$ with $|X| = \kappa$ and $|Y|\leq \kappa$.

Then $A_{X, Y} = \mathbb{Q}\times X\sqcup Y$ is a metric space with $|A_{X, Y}| = \kappa$ and $A_{X, Y}$ is homeomorphic with $A_{Z, W}$ (where $|Z| = \kappa$ and $|W|\leq \kappa$) if and only if $|Y| = |W|$. Since there is infinite amount of cardinals $\leq \kappa$ (for example, let $|Y| = 0, 1, 2, 3, ...$; in general, this gives us less than $\kappa$ examples, but always $\aleph_0$ many), we obtain an infinite amount of different metric spaces on a set of size $\kappa\geq \aleph_0$.

Note that there is total of $2^\kappa$ non-homeomorphic metric spaces of size $\kappa$ as per your other question.

Jakobian
  • 15,280
  • Why $A_{X,Y}$ is metric space? – amir bahadory Feb 16 '24 at 14:21
  • @amirbahadory Products and disjoint unions of a metric space can all be given a compatible metric i.e. they are metrizable. For proofs of these facts see here and here. So while I said its a metric space, I was being a little informal, and I meant that they can be given a metric. – Jakobian Feb 16 '24 at 14:33
  • Also, while the first link mentions they have to be bounded by $1$, if $d$ is a metric we can always take $\rho(x, y) = \min{d(x, y), 1}$ or $\rho(x, y) = \frac{d(x, y)}{1+d(x, y)}$ as another equivalent metric bounded by $1$. – Jakobian Feb 16 '24 at 14:37
  • I didn't want to give an explicit metric structure, because there is a lot of to choose from. For example, on the product there's also lots of choices of a metric e.g. $d((x, y), (t, z)) = (d_0(x, t)^p+d_1(y, z)^p)^{1/p}$ for any number $1\leq p < \infty$ gives a family of equivalent metrics on the product. – Jakobian Feb 16 '24 at 14:39