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

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Homework Help Overview

The problem involves determining the induced metric on the subspace of sequences consisting solely of zeros and ones within the space \( l^\infty \). The original poster attempts to explore the properties of the metric defined on \( l^\infty \) and how it translates to the subspace in question.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the metric defined on \( l^\infty \) and how it applies to sequences of zeros and ones. There is an exploration of specific cases where the values of the sequences are equal or different, and questions arise about the elegance and completeness of the proposed solution.

Discussion Status

Some participants have recognized the induced metric as the discrete metric, while others provide additional context regarding the implications of separability in metric spaces. The discussion appears to be productive, with various interpretations and insights being shared.

Contextual Notes

There is a mention of the importance of separability in functional analysis and its relation to the properties of the space \( l^\infty \). The implications of the metric's properties on the separability of the space are also under consideration.

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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.
 
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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.
 
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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.
 
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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.
 

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