Cardinality of a set of metrics

calculuscardinalsmetric-spacesreal-analysis

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 $p$?

Best Answer

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 $\mathbb{R} \to \mathbb{R}$.

Related Question