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Find the cardinality of the set of all metrics on $\mathbb{R}.$

The cardinality of the set of metrics on $\mathbb{R}$ is at most $|\mathbb{R}^{\mathbb{R}\times \mathbb{R}}|.$ Observe that $|\mathbb{R}^{\mathbb{R}\times \mathbb{R}}| = ((2^{\aleph_0})^{2^{\aleph_0}})^{2^{\aleph_0}}.$ Also, $|\mathbb{R}^\mathbb{R}| = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{2^{\aleph_0}}$ (indeed $2^{\aleph_0} \leq (\aleph_0)2^{\aleph_0}\leq 2^{\aleph_0}\cdot 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}$ so equality holds by the Cantor-Schroeder Bernstein Theorem). and so $|\mathbb{R}^{\mathbb{R}\times \mathbb{R}}| = (2^{2^{\aleph_0}\cdot 2^{\aleph_0}} = 2^{2^{\aleph_0 + \aleph_0}} = 2^{2^{\aleph_0}}).$ A metric on $\mathbb{R}$ is simply the standard metric, $d(x,y) := |x-y|.$ However, I'm not sure how to show that this cardinality is at least $2^{2^{\aleph_0}}.$

Edit: Initially, I believed the answer below was okay. However, after looking over the question again, I seem to have some problems.

  1. I can't seem to be able to prove that the cardinality of the set of metrics on $\mathbb{R}$ equals the cardinality of the set of metrics on $X$ (formally of course and using only basic cardinal arithmetic if possible).
  2. I tried to define a metric on $\mathbb{R}$ with $1$ taking the role of $p,$ but I couldn't really find an injective map (e.g. two such functions may differ at $1$). Perhaps I could consider continuous functions from $\mathbb{R}\to [1,2],$ but the cardinality of the set of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ is only $|\mathbb{R}|$.
  3. Or maybe I can make $\mathbb{R}\backslash \{0\}$ be treated as the "$\mathbb{R}$" in the answer by MikeF below and $\{0\}$ be treated as $p$?
user3472
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  • The first one probably requires some tricky insight. One approach is to observe that the interior of the unit disk union any subset of its boundary is convex (even more, any such set will be a star convex set). – Dave L. Renfro May 14 '21 at 15:51
  • For the second question, more than you'd probably want to know can be found by looking at the references I give in the comments to How many metric define on a nonempty set finite set or a countable set? – Dave L. Renfro May 14 '21 at 16:03
  • @spaceisdarkgreen sorry for the error. – user3472 May 15 '21 at 18:35
  • For (1) you can define $\phi:X \to \mathbb{R}$ by $\phi(p)=0$, $\phi(n)=n+1$ if $n$ is a nonnegative integer, and $\phi(x)$ if $x$ is a real number but not a nonnegative integer. You can check that's a bijection from $X\to \mathbb{R}$. Now, if $\phi:S_1\to S_2$ is any bijection and $d$ is a metric on $S_2$, you can define a corresponding metric $d_\phi$ on $S_1$ by $d_\phi(x,y)=d(\phi(x),\phi(y))$ for all $x,y \in S_1$. You can check that $d \mapsto d_\phi$ defines a bijection from the set of metrics on $S_2$ to the set of metrics on $S_1$. – Mike F May 29 '21 at 04:13
  • About (2), in the answer I gave below it is not necessary for the function $f : \mathbb{R} \to [1,2]$ to be continuous. It can just be an arbitrary function. – Mike F May 29 '21 at 04:24
  • About (3), yes this makes sense. Given any function $f : \mathbb{R} \setminus {0} \to [1,2]$ (it does not need to be continuous), you can define a metric $d_f$ on $\mathbb{R}$ by putting $d(x,y)=1$ if $x,y \in \mathbb{R}\setminus{0}$ and $x \neq y$, putting $d(x,0)=d(0,x)=f(x)$ for $x \in \mathbb{R}\setminus{0}$, and putting $d(x,x)=0$ for all $x \in \mathbb{R}$. There are as many such metrics as there are functions $\mathbb{R}\setminus {0} \to [0,1]$. – Mike F May 29 '21 at 04:28

1 Answers1

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For convenience, let's look at metrics on the underlying set $X = \mathbb{R} \cup \{p\}$, where $p$ is some point not in $\mathbb{R}$. Then, any function $f :\mathbb{R} \to [1,2]$, determines a unique metric $d_f$ on $X$ such that:

  1. the restriction of $d_f$ to $\mathbb{R}$ is usual discrete $\{0,1\}$-valued metric on $\mathbb{R}$,
  2. $d_f(p,x)=f(x)$ for all $x \in \mathbb{R}$.

From this construction, you can deduce that there are at least as many metrics on $\mathbb{R}$ as there are functions $\mathbb{R} \to \mathbb{R}$.

Mike F
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  • I don't understand your solution. Why construct a metric on $X\neq \mathbb{R}$ when you want to find a lower bound for the cardinality of the set of metrics on $\mathbb{R}$? Also, can $f$ be any function from $\mathbb{R}$ to $[1,\infty)$ and should $d_f(p,p)=0$? – user3472 May 15 '21 at 18:43
  • Yes it's true that $|\mathbb{R}^\mathbb{R}|=2^{2^{\aleph_0}}$ and your argument seems OK to me. You could also look here: https://math.stackexchange.com/questions/17914/cardinality-of-the-set-of-all-real-functions-of-real-variable. – Mike F May 15 '21 at 18:48
  • I stipulated that $f$ takes values in $[1,2]$ so that the triangle inequality would be satisfied. There are some other restrictions on $f$ which would work for that purpose, but this just seemed like an easy one. I don't think you get a metric for any function $f : \mathbb{R} \to [1,\infty)$. You can't be sure that $f(x) = d(p,x) \leq d(p,y)+d(y,x) = f(y)+1$ for all $x,y\in\mathbb{R}$. – Mike F May 15 '21 at 18:54
  • As for why use this space $X$, you realize that $X$, $[1,2]$, $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ all have the same cardinality? – Mike F May 15 '21 at 18:56
  • @user3472: Yes for sure you can do that. I started out writing it that way, but then decided to introduce an extra point $p$ for cosmetic reasons. – Mike F May 17 '21 at 16:02