# Finding the Induced Metric

1. May 31, 2015

### Bashyboy

1. The problem statement, all variables and given/known data
If $A$ is the subspace of $l^\infty$ consisting of all sequences of zeros and ones,
what is the induced metric on $A$?

2. Relevant equations

3. 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.

2. May 31, 2015

### pasmith

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 $x_i \neq y_i$.

3. Jun 3, 2015

### micromass

Staff Emeritus
This metric has a special name that you might have seen. Do you know what it is?

4. Jun 3, 2015

### Bashyboy

Yes, I do indeed know: it is the discrete metric!

5. Jun 3, 2015

### micromass

Staff Emeritus
Exactly! Well done.

6. Jun 3, 2015

### micromass

Staff Emeritus
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.