Understanding Countable Sets in Measure Zero Definition

[gotjrgkr]

Homework Statement

While studying a book "analysis on manifolds" by munkres, I see a definition of measure zero. That is,
Let A be a subset of R^{n}. We say A has measure zero in R^{n} if for every ε>0, there is a covering Q_{1},Q_{2},... of A by countably many rectangles such that \sum_{i=1}^{\infty}v(Q_{i})<ε.

But, in this text, there's no remark about the countable set. I mean, it seems to me that countable set is not defined.

Could you tell me what the "countably many" mean in the above definition??

Homework Equations

The Attempt at a Solution

 

[CompuChip]

A countable set is a set from which you can find an injection to the naturals.

 

[HallsofIvy]

Munkres assumes that you already know what "countable" means. As CompuChip said, a set is countable if there is a one to one mapping to a subset of the positive integers. (Some textbooks use the term "countable" to mean an infinite set that can be mapped to a subset of the positive integers, some use the term to include finite sets.)

 

[gotjrgkr]

Ah,
I really thank you for your reply.
But, I want to check myself if I understand your reply well.

If I use the term "countable_#" for a set A to indicate that there's a bijection from the set A to the set of all natural numbers, the word "countable" in the textbook would
mean either "finite" or "countable_#", right?? Am I right?


But, I still have a difficulty to understand the meaning of the definition of measure zero.
I don't have any idea how to see the case when for a given ε>0, there's a finite set of rectangles covering a given set. what does the \sum_{i=1}^{\infty}v(Q_{i}) mean in this case??

 

[HallsofIvy]

I presume that the Q_i are non-overlapping sets whose union is some larger set, A, say. That says that the measure of the set A is the sum of the measures of each individual Q_i. Notice that they are indexed by i going from 1 to infinity. That is, the sets are labeled Q_1, Q_2, ... That labeling itself maps the sets to the positive integers so there must be a "countable" number of such sets (which are not necessarily countable themselves).

 

[gotjrgkr]

I presume that the Q_i are non-overlapping sets whose union is some larger set, A, say. That says that the measure of the set A is the sum of the measures of each individual Q_i. Notice that they are indexed by i going from 1 to infinity. That is, the sets are labeled Q_1, Q_2, ... That labeling itself maps the sets to the positive integers so there must be a "countable" number of such sets (which are not necessarily countable themselves).

I'm a little confused...
If a set A has measure zero, then
do you mean that the meaning of "countable" in the text is such that for any given ε>0, there is a set of rectangles Q_{i} which is mapped to the set of all natural numbers by a bijection(I mean, this bijection is a one to one mapping of the set of rectagles Q_{i} onto the set of all natural numbers) and whose sum of the volumes of rectangles is less than ε and union of the rectangles contains A?

 

[CompuChip]

Maybe an explicit example will help here.

Suppose you take the line segment from (0, 0) to (1, 0) in R² and pick a value for ε.
If you take a rectangle with corners (0, -ε/2), (0, ε/2), (1, ε/2) and (1, -ε/2) you have a set with one rectangle that covers the line segment and has total measure ε. This is a very easy way to demonstrate that the line segment has measure zero. If you insist that "countable" excludes "finite" you can of course get the same result but through a longer route :-)

If you consider the x-axis in R², you can no longer do with finitely many rectangles even if you wanted.
What you can do, for example, is let Q_n be a rectangle around each integer, of height ε / n² (I think - you might want to check this). There will be infinitely many, but still countably many.

 

[gotjrgkr]

Maybe an explicit example will help here.

Suppose you take the line segment from (0, 0) to (1, 0) in R² and pick a value for ε.
If you take a rectangle with corners (0, -ε/2), (0, ε/2), (1, ε/2) and (1, -ε/2) you have a set with one rectangle that covers the line segment and has total measure ε. This is a very easy way to demonstrate that the line segment has measure zero. If you insist that "countable" excludes "finite" you can of course get the same result but through a longer route :-)

If you consider the x-axis in R², you can no longer do with finitely many rectangles even if you wanted.
What you can do, for example, is let Q_n be a rectangle around each integer, of height ε / n² (I think - you might want to check this). There will be infinitely many, but still countably many.

So, in the first case, there's only one rectangle for the given ε, so that it's finite and it implies there is a finite set(that is, countable set) of rectangles(actually only one) which covers the line segment, right?
I can imagine what you say about the first case, but could you explain more specifically about the second case?

 

[CompuChip]

OK.
What I meant is, let Q₀ be the rectangle (-1/2, 1/2) x (-ε'/2, ε'/2). I use the normal Cartesian product notation, so
(a, b) \times (c, d) = \{ (x, y) \mid x \in (a, b), y \in (c, d)
and (a, b) is the open interval a < x < b.

For all other integers n, define
Q_n = (n - 1/2, n + 1/2) x (-ε' / 2n², ε' / 2n²).

Now you can compute the total surface:
\sum_{n = -\infty}^{\infty} \nu(Q_n) = \epsilon&#039; + 2 \sum_{n = 1}^\infty \frac{\epsilon&#039;}{n^2}
(this is straightforward, although it takes some sum manipulations to get there).

Using a standard result that
\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
you can show that you get \epsilon&#039;(1 + \pi^2 / 3).

Some things I'll leave for you to think about:
(*) I have defined Q_n for 0 and negative integers as well. However, the set of all these rectangles is still countable (hint: 0, 1, -1, 2, -2, ...).
(*) If you scale \epsilon&#039; in the preceding by a suitable constant (e.g. \epsilon = \epsilon&#039; / (1 + \pi^2 / 3) should do) you can get the result smaller than \epsilon.
(*) You might want to run through my calculations - I have (intentionally) skipped some of the more technical steps. Let me know if you can't follow.

 

[gotjrgkr]

OK.
What I meant is, let Q₀ be the rectangle (-1/2, 1/2) x (-ε'/2, ε'/2). I use the normal Cartesian product notation, so
(a, b) \times (c, d) = \{ (x, y) \mid x \in (a, b), y \in (c, d)
and (a, b) is the open interval a < x < b.

For all other integers n, define
Q_n = (n - 1/2, n + 1/2) x (-ε' / 2n², ε' / 2n²).

Now you can compute the total surface:
\sum_{n = -\infty}^{\infty} \nu(Q_n) = \epsilon&#039; + 2 \sum_{n = 1}^\infty \frac{\epsilon&#039;}{n^2}
(this is straightforward, although it takes some sum manipulations to get there).

Using a standard result that
\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
you can show that you get \epsilon&#039;(1 + \pi^2 / 3).

Some things I'll leave for you to think about:
(*) I have defined Q_n for 0 and negative integers as well. However, the set of all these rectangles is still countable (hint: 0, 1, -1, 2, -2, ...).
(*) If you scale \epsilon&#039; in the preceding by a suitable constant (e.g. \epsilon = \epsilon&#039; / (1 + \pi^2 / 3) should do) you can get the result smaller than \epsilon.
(*) You might want to run through my calculations - I have (intentionally) skipped some of the more technical steps. Let me know if you can't follow.

I got it! I see what you mean..
Anyway, in the munkres's text, he also treats a finite set as countable set as well as an infinite set mapping to the set of positive integers by one-to-one and onto fashion.
I was confused because in some other books, for example, rudin's principle of mathematical analysis, a set A is said to be countable if there is a bijection between A and the set of natural numbers.
I wanted to know which one is used in the munkres's text.
I really appretiate all of you. It helps me a lot!
(please, let me know if there is a wrong part in the statement above that I wrote.)