What is the Induced Metric on the Subspace of Zeros and Ones in ##l^\infty##?

  • Thread starter Thread starter Bashyboy
  • Start date Start date
  • Tags Tags
    Induced Metric
Bashyboy
Messages
1,419
Reaction score
5

Homework Statement


If ##A## is the subspace of ##l^\infty## consisting of all sequences of zeros and ones,
what is the induced metric on ##A##?

Homework Equations

The Attempt at a Solution


The metric imposed on ##l^\infty## is ##d(x,y) = \underset{i \in \mathbb{N}}{\sup} |x_i - y_i|##. I suspect that the induced (or, as a I call it, the reduced) metric is ##d(x,x) = 0## and ##d(x,y) = 1##. However, I am having difficulty showing this. Here is what I came up with:

Let ##I_{x,0} = \{i : x_i = 0\}## and ##I_{x,1} = \{i : x_i = 1\}##, and similarly define ##I_{y,0}## and ##I_{y,1}##. Hence, ##I_{x,1} \cap I_{y,1} \ne \emptyset## implies that there exists an ##i## such that ##x_i = 1## and ##y_ i = 1##; furthermore, ##x_i - y_i = 0##, which means ##\underset{i \in \mathbb{N}}{\sup} |x_i - y_i| = 0##...right?

This, however, does not appear to be a very elegant solution, as there will be many cases to deal with.
 
Physics news on Phys.org
If x_i and y_i take values in \{0,1\} then the only possible values of |x_i - y_i| are 0 if x_i = y_i and 1 if x_i \neq y_i.

If x = y then by definition x_i = y_i for all i \in \mathbb{N}.
If x \neq y then by definition there exists an i \in \mathbb{N} such that x_i \neq y_i.
 
  • Like
Likes Bashyboy
This metric has a special name that you might have seen. Do you know what it is?
 
Yes, I do indeed know: it is the discrete metric!
 
Exactly! Well done.
 
  • Like
Likes Bashyboy
Some further information on why this problem is important: A metric space is separable if (by definition) it has a countable dense subset. Separability is an important condition in functional analysis and (among others) it ensures that a Hilbert space has a countable orthonormal basis. Now you can show that a discrete metric space is separable iff it is countable. So your ##A## is not separable. This implies immediately that ##\ell^\infty## is not separable since subspaces of separable spaces are separable (this is not that easy to show). ##\ell^\infty## is one of the most important spaces that is nonseparable. The other ##\ell^p## spaces are separable.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top