What metrics can be used on the power set of the reals?

Question

Let $2^\textbf{R}$ be the power set of $\textbf{R}$, that is, the set of all subsets of the real numbers. What metrics can I use to turn $2^\textbf{R}$ into a metric space?

Of course we have the discrete metric, but I'm curious if a metric exists that gives some kind of notion of distance between two sets. The Hausdorff distance function almost does this, but it isn't a metric, because the distance between two different sets can be 0.

What metrics are there that turn $2^\textbf{R}$ into a metric space? Also curiously, could any of these metrics be extended to $2^{(2^\textbf{R})}$, the power set of $2^\textbf{R}$, as well as larger power sets?

Answer 1

There are loads of metrics!

Just choose your favourite metric space $X$ of cardinality $2^{\frak c}$ and give $2^\mathbb R$ the metric inherited from your favourite bijection $X \to 2^\mathbb R$.

One problem with this approach is the subset $\mathbf R =\{\{x\} : x \in \mathbb R\}$ might not be homeomorphic to $\mathbb R$ -- which allows you imagine $\mathbb R$ sitting inside the space $2^\mathbb R$. For example if the bigger space is discrete so is the smaller space.

Okay we'll demand $\mathbf R \cong \mathbb R$ but now what?

Just choose your favourite metric space $X$ of cardinality $2^{\frak c}$ and bijection $X \to (2^ \mathbb R - \mathbf R)$. Use this to put a metric $d'$ on $(2^ \mathbb R - \mathbf R)$. Give $\mathbf R$ the metric $\rho'$ to make it into the real line. Then replace each of the two metrics with their truncation.

$d(x,y) = \displaystyle \frac{d'(x,y)}{1+d'(x,y)}$

$\rho(x,y) = \displaystyle \frac{\rho'(x,y)}{1+\rho'(x,y)}$

This ensures all distances are less than $1$. Then put $(2^ \mathbb R - \mathbf R)$ and $\mathbf R$ back together by defining any $x \in (2^ \mathbb R - \mathbf R)$ and $y \in \mathbf R$ to be distance $2$ apart.

The resulting space is just a copy of $\mathbb R$ along with some arbitrary metric space beside it. It doesn't really capture how the other elements are supposed to be subsets of the real line. For that we have to make further demands.

I think a suitable demand is to specify what you want the metric to look like on the subspace $\mathcal F (\mathbb R) = \{A \subset \mathbb R : A$ is finite$\}$ and try to expand from there. We might want for example want the Hausdorff metric:

$d(A,B) = \min \{\varepsilon > 0 :$ each $a \in A$ is within distance $\varepsilon$ of some $ b \in B$ and vice-versa$\}$

Now let me show you what happens when we make the semi-reasonable demand that $\mathcal F (\mathbb R)$ is dense.

Since $\mathbb Q$ is dense in $\mathbb R$ it can be shown that under the Hausdorff metric $\mathcal F (\mathbb Q) = \{A \subset \mathbb Q : A$ is finite$\}$ is dense in $F (\mathbb R)$. By extension $F (\mathbb Q) $ is dense in the whole space.

However the subspace $F (\mathbb Q) $ has cardinality $|\mathbb Q| = |\mathbb N|$ This shows $2^{\mathbb R}$ must be is separable. And unfortunately every separable metric space has cardinality at most $2^{|\mathbb N|}$. This shows $2^{\mathbb R}$ is simply too large to admit an extension of the Hausdorff metric such that $F (\mathbb R)$ is dense.

So we need to cook up less restrictive conditions. . . .

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Edit: Here's an example:

The space $\mathbb (0,1) ^ \mathbb R$ of all functions $(-1,1)\to \mathbb R$ is a metric space under the metric:

$d(f,g) = \displaystyle \max \{|f(x)-g(x)| : x \in (0,1)\}$

But what is the cardinality? To calculate that first observe that $|\mathbb R| = |\mathbb N||\mathbb R|$ and $|\mathbb R| = |(0,1)|$. Then compute. . .

$\displaystyle 2 ^ {|\mathbb R|} = 2 ^ {|\mathbb N||\mathbb R|} = (2 ^{|\mathbb N|} )^{|\mathbb R|} = |\mathbb R|^{|\mathbb R|} = |(-1,1)|^{|\mathbb R|} = |(-1,1)^{\mathbb R}|$

So this space has the correct cardinality