# Fun with metrics

• Sep 23rd 2011, 10:11 PM
SlipEternal
Fun with metrics
I'm a first year grad student. After listening to my algebra professor's explanation that for any finite set A, P(A) is a vector space under the operation of symmetric difference over the field $\displaystyle \mathbf{Z}_2$. It got me to wondering what a metrization of the power set might look like. So, I came up with:

$\displaystyle d: P(A)\times P(A) \to $0,1$$
$\displaystyle d(U,V)=\frac{card\{U\triangle V\}}{card\{U\bigcup V\}}$
Where d is defined to be zero if both U and V are the empty set, and I haven't formulated anything for infinite sets, yet.

This seems like it would, indeed, be a metric on P(A), but it induces the discrete topology, obviously. (Well, it wasn't obvious to me, but it was to all of my professors who found my efforts entertaining until I discovered that I was attempting to prove the existence of a metric topologically equivalent to the 0-1 metric). I haven't actually proven the triangle inequality yet, but it seems likely that I could if I expended the effort.

Anyway, it wasn't a total waste of time. I learned quite a bit from the exercise. And I would like to continue my efforts. So, I'm reevaluating my strategy. I thought I would start with something a little simpler, which I know more about, but for which my professors don't believe a metric can be found which has a topology more interesting than the discrete one.

Rather than an arbitrary power set, I thought I would try:
Let $\displaystyle X=\{A \subseteq \mathbf{Z} | A \text{ is finite}\}$
Now, I have a lot more things I can try to play around with. For instance, some form of sparseness function that calculates maximal element - minimal elemnt and divides by cardinality? I haven't decided yet how it might work, but that seems to work under unions (where the sparseness of the union would be less than or equal to the sum of the sparseness for the separate sets). I made up the word sparseness, of course, but I haven't thought about it enough to give it a decent term.

Anyway, any ideas for how I might proceed would be greatly appreciated! Does anyone know if I can norm a power set in such a way that it can be metrized with the induced topology anything other than discrete?
• Sep 23rd 2011, 11:01 PM
Drexel28
Re: Fun with metrics
Quote:

Originally Posted by SlipEternal
I'm a first year grad student. After listening to my algebra professor's explanation that for any finite set A, P(A) is a vector space under the operation of symmetric difference over the field $\displaystyle \mathbf{Z}_2$. It got me to wondering what a metrization of the power set might look like. So, I came up with:

$\displaystyle d: P(A)\times P(A) \to $0,1$$
$\displaystyle d(U,V)=\frac{card\{U\triangle V\}}{card\{U\bigcup V\}}$
Where d is defined to be zero if both U and V are the empty set, and I haven't formulated anything for infinite sets, yet.

This seems like it would, indeed, be a metric on P(A), but it induces the discrete topology, obviously. (Well, it wasn't obvious to me, but it was to all of my professors who found my efforts entertaining until I discovered that I was attempting to prove the existence of a metric topologically equivalent to the 0-1 metric). I haven't actually proven the triangle inequality yet, but it seems likely that I could if I expended the effort.

Anyway, it wasn't a total waste of time. I learned quite a bit from the exercise. And I would like to continue my efforts. So, I'm reevaluating my strategy. I thought I would start with something a little simpler, which I know more about, but for which my professors don't believe a metric can be found which has a topology more interesting than the discrete one.

Rather than an arbitrary power set, I thought I would try:
Let $\displaystyle X=\{A \subseteq \mathbf{Z} | A \text{ is finite}\}$
Now, I have a lot more things I can try to play around with. For instance, some form of sparseness function that calculates maximal element - minimal elemnt and divides by cardinality? I haven't decided yet how it might work, but that seems to work under unions (where the sparseness of the union would be less than or equal to the sum of the sparseness for the separate sets). I made up the word sparseness, of course, but I haven't thought about it enough to give it a decent term.

Anyway, any ideas for how I might proceed would be greatly appreciated! Does anyone know if I can norm a power set in such a way that it can be metrized with the induced topology anything other than discrete?

Well, your effort do metrize $\displaystyle 2^A$ beyond the discrete metric is doomed. Indeed, if you take $\displaystyle A$ finite then any $\displaystyle T_1$ (points closed) space will be discrete, right? This is because any subset of $\displaystyle A$ can be written as a union of finitely many (since $\displaystyle A$ itself is finite) singeltons and so is closed. Thus, every subset of $\displaystyle A$ is closed and so every subset of $\displaystyle A$ is open. The only metric which induces a discrete topology can be taken to be the discrete metric, so...well, that's it. You're doomed.

Now, for $\displaystyle X$ it depends how precisely you define 'interesting'. If you mean just 'cool' you could recall (a proof can be found on my blog here) that the set of all finite subsets of an infinite set is equipotent to the set itself. Thus, $\displaystyle X$ is countably infinite. In particular, we can find a bijection $\displaystyle f:\mathbb{Q}\to X$. We can then take the $\displaystyle p$-adic metric $\displaystyle d_p$ on $\displaystyle \mathbb{Q}$ and define a metric $\displaystyle d$ on $\displaystyle X$ to be exactly such that $\displaystyle f: (\mathbb{Q},d_p)\to (X,d)$ is an isometry. You could even do the same thing but giving $\displaystyle \mathbb{Q}$ the usual metric. Does that work, or do you want something more natural?
• Sep 23rd 2011, 11:12 PM
Drexel28
Re: Fun with metrics
Let me add, and I haven't thought too much about this, but I may have a way to make 'interesting' more precise. Namely, given an infinite set $\displaystyle X$ can we find a 'natural' (something not too contrived) metric on $\displaystyle P(X)$ that turns it into a metric group with it's standard operations. Note, that since the constant multiplication maps $\displaystyle 0:P(X)\to P(X):x\mapsto 0x$ and $\displaystyle 1:P(X)\to P(X):x\mapst 1x$ are just a constant map and the identity map respectively, any metric which turns $\displaystyle P(X)$ into a metric group will consequently turn it into a topological $\displaystyle \mathbb{Z}_2$-vector space.
• Sep 24th 2011, 06:04 AM
SlipEternal
Re: Fun with metrics
For interesting, currently, I'm starting with cool, and working towards 'natural'. Pretty much, this exercise is supposed to help me understand metrics better, as well as methods of research (one of my professors recommended that I post here about it). I'm really glad that he did, because I hadn't heard of the P-adic metric before. Very interesting!

Thank you again. I have a busy weekend ahead of me now :D
• Sep 24th 2011, 07:05 AM
SlipEternal
Re: Fun with metrics
So, any bijection $\displaystyle f: \mathbb{R} \to \mathcal{P}(\mathbb{Z})$ would allow the metric $\displaystyle \mathfrak{d}$ to be defined exactly so that $\displaystyle f: (\mathbb{R},d) \to (\mathcal{P}(\mathbb{Z}),\mathfrak{d})$ is an isometry? That's pretty straightforward, but probably not terribly 'natural' in the sense that most such bijections would yield nonsense distances. Like, given $\displaystyle A \subseteq \mathbb{Z}$, define the sequence $\displaystyle a_n=1$ if $\displaystyle (-1)^n\lfloor \frac{n}{2}\rfloor \in A$ and zero otherwise $\displaystyle (n \in \mathbb{N})$. I don't know what to do with sequences that end in all 1s, but otherwise, it is bijective with the binary expansion of decimals in the $\displaystyle $0,1$$ interval, right?