# What is zero content?

• Apr 12th 2011, 06:41 AM
calculuskid1
What is zero content?
I know the definition, But I'm looking for an explanation.. anyone?
• Apr 12th 2011, 06:47 AM
Plato
Quote:

Originally Posted by calculuskid1
I know the definition, But I'm looking for an explanation.. anyone?

How about telling us the definition that you do have?
There are variations on similar ideas.
• Apr 12th 2011, 07:21 AM
calculuskid1
I have:
for every epsilon>0 there is a finite collection of intervals I1...IL st:
A) Z⊂U IL,and
B) The sum of the lengths of the I l's is less than epsilon.
But how can the sum of the intervals be less than epsilon, does it just mean that zero content are intervals that are arbitrary small..?
• Apr 12th 2011, 07:53 AM
Opalg
Quote:

Originally Posted by calculuskid1
I have:
for every epsilon>0 there is a finite collection of intervals I1...IL st:
A) Z⊂U IL,and
B) The sum of the lengths of the I l's is less than epsilon.
But how can the sum of the intervals be less than epsilon, does it just mean that zero content are intervals that are arbitrary small..?

For a start, any finite set of points has zero content. Given a set with n points, you can surround each of them by an interval of length $\varepsilon/n$.

For an example of an infinite set with zero content, let $S = \{1/n:n=1,2,3,\ldots\}$. Given $\varepsilon>0$, the interval $(0,\varepsilon/2)$ contains all but finitely many elements of S, and you can use the remaining $\varepsilon/2$ of length to put little intervals around those finitely many points.
• Apr 12th 2011, 08:01 AM
Plato
Quote:

Originally Posted by calculuskid1
I have:
for every epsilon>0 there is a finite collection of intervals I1...IL st:
A) Z⊂U IL,and
B) The sum of the lengths of the I l's is less than epsilon.
But how can the sum of the intervals be less than epsilon, does it just mean that zero content are intervals that are arbitrary small..?

Suppose that $\varepsilon > 0$ the open interval $O_n = \left( {a - \frac{\varepsilon }{{2^{n + 1} }},a + \frac{\varepsilon }{{2^{n + 1} }}} \right)$ has length $\ell (O_n ) = \frac{\varepsilon }{{2^n }}$.
The sum of all those is just $\sum\limits_{n = 1}^\infty {\ell (O_n )} = \varepsilon$.
Thus $\{a\}$ has content zero.

Now that is a simple-minded example that has a natural extension to larger sets.
Essentially the set has ‘zero area’. Think about any open interval say $[a,b]$.
If $\varepsilon = \frac{{b - a}}{2} > 0$ it would be impossible to cover $[a,b]$ with set whose total length is less that $\varepsilon$.

Edit: I did not see reply #4 before posting, sorry.
• Apr 13th 2011, 10:59 AM
Deveno
content is supposed to correspond intuitively to length (in 1 dimension), area (in 2 dimensions) or volume (in higher dimensions).

i say "supposed" because it turns out that historically, this has been a hard notion to pin down. the problem isn't with "normal" things like lines, triangles, circles, squares, etc. it has to do with the fact that there are lots and lots of kinds of sets, and some of them are pretty weird.

consider this set: {x in [0,1] : x is rational}. what is the "length" of this set? how would we even begin to measure it? it turns out that different notions of measurement, give different answers.