The Heine-Borel Theorem, 3

                      ←  Part II  

{CLEANUP TO FOLLOW}

I swear it's got to be in here somewhere.  I think it's safe to move on now.

I say:

INTERVAL WITH OPEN END AT INFINITY

(1) The Postulate of Eudoxus:

If I say, "n is the greatest number", then you say

"n plus one", and I lose.

As stated by Euclid, the postulate itself is *cough* a series of interleaved ratios.

(2) Bob Hughes' Fishing Lemma:

"I went fishing and caught a fish this big."  He shows the size with a gesture.

Bob Hughes has one arm.

INTERVAL WITH END AT ZERO

(3) Postulate of Eudoxus:

If I say, "p is the closest point to c," then

You say, "p minus c over two",  and I lose again.

(4) Imagine a Film Reel of a woman throwing a ball. The frames are time-stamped.  Call the time on the first frame a, and the last, b.

• (a) Say the recording begins the instant the ball leaves her hand.
  When did the ball leave her hand?      Go to the first frame:     t = a. 
• (b) Instead, say we must find the moment the ball leaves her hand.  We find the last frame where the ball is still touching her hand.  In the next, there is sky between her fingers.  The moment lies between these two frames.
  Suppose the recording is very fast:  there are as many frames per second as we care to examine. You start on the left where there is contact and  I begin on the right, with sky, and we move toward one another.     We can make the time between the two adjacent frames   --touching the ball, then sky-- as narrow as we want.  We can make it narrower than any given number, no matter how small.
  
  When did the ball leave her hand?
  We may answer in several ways.  Any finite number is an approximation.  But we can make our error vanish.  I suggest the following way to answer:
• (c) Suppose at time t₀ there is a frame which appears to be the last moment the ball is touching her hand.  In every frame which follows, there is sky between her fingers and the ball.  For convenience, cut the film so this frame is first and label it t=0.
    Consider a small strip of film from t=0 to  to t = δ.   δ is very small, so we pick the halfway point and say, "the ball leaves her hand at t = δ/2".  But when we check, there is already sky at t = δ/2.   And at t =  δ/4,  δ/8.... δ/n.  We can find no n large enough to give us another image of the ball in contact with her hand.
  The ball is touching her hand at t=0.  But the first frame with sky does not exist.  We can name no number small enough to identify it.  Measuring from the right, the first frame of sky is lim_{{δ → 0}}δ = 0.  And the instant we are looking for, squeezed between the two zeros!
  
  Then we can answer, When did the ball leave her hand?   with   "t = 0."
  And we have defined a way to build an interval which "begin at the instant.....".

I suspect the difference between (b) and (c) is the crux of the Heine-Borel Theorem.
Note that we must begin with the assumption of the limit point in order to measure this way.  In (b) the limit point is not known, and cannot necessarily be identified.  The procedure has no algebraic limit;  the limit remains a question of time, and our willingness to look at film.  For example, a clever rule which gives us alternating frames of sky and contact might at first seem good enough... or perhaps it closes on a nearby point instead, and we have not let it narrow in closely enough for the frames to stop alternating.  Our rule was a guess just like any other.
  In Widder's example of the open interval at 0, the limit point is known and the dilemma is pretend: measuring without the endpoint gives the algebraic limit anyway, and so we cut the first frame.
The behavior of a function near limit points is a different question.

(5) Define COVER:

I have a line segment and some opaque junk.  Let's pile some junk on top of the line segment. I say,

Definition:  the line segment is covered in junk if no point of the segment is visible. (there is at least one piece of junk on top of every point of the segment)

Definition:  the line segment is exactly covered if

     *no point is visible and

     *no point beyond the endpoints of the segment is covered by junk.

That is, [a,b],  and only [a,b], is covered in junk.

The closed interval can be exactly covered by a finite number of subintervals.  We can divide it up however we want; it is a whole measure.    It is also possible to exactly cover a whole measure using an infinite series of bits of junk.  Just divide in a way that always shrinks the distance between, but also never arrives.  In other words, just divide, period.

MISCELLANY

(6)  The Obvious:  It is impossible to measure the exact length of an open interval because, by definition,  it doesn't have one.   There is no left endpoint of (a, b].

If we remove the limit point, we can't get to it by division any more than by offset, because it's not there.

Division's nice because it keeps me from popping out on the other side.

(7) The Magician

A magician asks you to draw a card and look at it.  What's my card?  She says,  "The card you drew is the one you're holding!"

• Choose any number, and don't tell me which one.  Let's call it r.  How far is r from the origin?  Why," r is r units from the origin."  

• Now, choose two positive integers.  Call the smaller one p and the larger, q.  Consider the number     α = q/(q − p)How far is α from the origin?     q/(q − p) units.

• Now, using the same p and q as above, let β =  This limit is q/(q − p).  So how far is β from the origin?  q/(q− p) units.  So α = β.

 The same number obtained in different ways.  The limit exists if we agree that nothing can stop us from adding more terms.  

Neither finite division nor continuity are the consequence of infinite division.  Buying a really fast clock and watching the digits spin by guarantees that I can get from one second to the next?  I get there if I get there.  You check.  Should I just skip ahead to limit tests?

Continuity exists is we agree that nothing can stop us from dividing.  We agree to measure continuously.  No argument can force physical objects to skip from place to place if they do not. No argument can prevent me from putting any number howsoever chosen, into the equation y = f(x).
So  f(x) is continuous.  
Unless, you know.... it isn't.
Get it?  The determining factor is not the property of the interval itself.  Proceeding by proof at this point is stupid. This is the time to go by definition.  Define the tools we need, to examine functions and retain the measurable characteristics.

(8)Slide the Subinterval:
Take a subinterval I' of some finite width.  Slide it all the way to the left.  Stop when the left edge of the subinterval covers the endpoint.

Suppose the left end of the interval  is open.
   Good luck with that.
   You can get closer and closer to the edge, but no matter where you stop, there is more interval to your left.  Don't overshoot.  The task is impossible.

Suppose the left end of the interval is closed.
   *slide*...  *click*.  The subinterval snaps into place when the left edge hits a.

Now we can continue: Holding a fixed, shrink I' to zero.  Now, scale (*cough*) it back up until the right edge coincides with b.  Holding b fixed, shrink I' to zero.

The test which fails for the open interval is offset.  Once this test fails, we cannot resolve it by division, or we could have done it with addition in the first place.  

It may seem as if division "keeps" me from popping out on the other side, but that is an illusion.  It only does so if I already know the limit, and measure my divisions from it.

(9) Diagrams:

Begin with a closed interval [a, b] and call it I.

I would like to prove that we can cover the whole interval [a, b] with a finite number of subintervals.  The center of each subinterval must be in [a,b].

Select  n points on the interval and  label them in ascending order:

We may choose   c1 = a   and/or   c2 = b,  but we do not have to.

 I'll chose one of my points at random, and call it c. Regular-ass c.  It can be any of the c_i.  In the picture above, see how the points are shown as filled circles. Zoom in on one of these circles.  It represents a point on the line, at the center of the circle.  Call the center C and the radius δ_c:

Q: Is there more than one point of the number line inside the circle?

A: Yes.    There are infinitely many as long as the radius is not zero.  As long as the radius is not zero.  For suppose I claim there is some δ_c > 0 which is the closest point to c, so that only c lives inside the circle.

Then you name ε = δ_c/2, and 0 < ε < δ_c, and I lose.

We can play this game forever.

So we can cover an arbitrary segment of the number line with a circle.  Ignoring the y-axis, an interval  [c−δ_c,  c+δ_c], with center at c.

Q: So what?

A: If we can guarantee that every point c in the whole interval a ≤ c ≤ b is the center of such a circle, then the rest follows.  Because every point is also then inside the circle of some other point.  Whatever the distance is between these two points, call it η.  Then we can cover the whole interval a ≤ c ≤ b with  (b− a)/η subintervals.

Again, we could just agree to divide and be done with it. All the rest ... ALL OF IT, is unnecessary.

So what is the Heine-Borel theorem?

 It seems that

•  It must relate several points:   *Different functions of c behave differently in the absence of an endpoint.   *Division is bounded from below by zero, but multiplication is not bounded from above.For example, in all the cases above, the interval is finite, the lower bound is zero for both a finite and infinite number of I_k, for both closed and open endpoint.  The same cannot be said for an endpoint open at infinity.
• . . .Where we want . . .   *δ_c to be a function of c.   *To use the same definition when the interval involves zero.
      *To distinguish limits with intervals.

I am left with the question,
What is a correct statement of the Heine-Borel Theorem?

The Heine-Borel Theorem, 2

←  Part I                                           Part III →

Reiterating, we have from David Widder's Advanced Calculus:

" Let there correspond to each point c of the closed interval a ≤ x ≤ b a number δ_c and an interval I_c of length 2δ_c with c the center point,
 .

The Heine-Borel theorem states that a finite number of the intervals (1) can be chosen which will cover the whole interval a ≤ x ≤ b.  That is, every point of a ≤ x ≤ b will be in at least one of the above mentioned finite number of intervals (1). In order to emphasize the need for proving this result, let us give an example to show it false if the interva, (a,, b) were open instead of closed.
   Let a = 0,  b = 1, and define I_c, for 0 < c < 1, as

That is, δ_c = c/2. No finite set of the intervals (2) will cover the interval 0 < x < 1.  For, consider such a set, I_c1, I_c2, . . . , I_cn, where 0 < c₁ < c₂ < . . . < c_n.  Of these, the interval I_c1 reaches farthest to the left.  Hence no point to the left of c₁/2 is covered by the set.  "   (pg. 170)

Now

 "Theorem 5:  To each c,  a ≤ c ≤ b corresponds an interval (1)  There exist points  c₁, c_2,, . . ., c_n of a ≤ x ≤ b, such that every point of the interval a ≤ x ≤ b is in at least one of the intervals I_c1, I_c2, . . . , I_cn.

Call a point A of the interval I, a ≤ x ≤ b, accessible if the interval a ≤ x ≤ A can be covered by a finite sequence of the intervals I_c.  Clearly, if A is accessible, every point of I to the left is also.  Hence, there must either be a point B of I dividing accessible points from inaccessible ones, or else all points of I are accessible.  (Some points are accessible, since all points of  I in I_a are covered by the single interval I_a.)  But the existence of the dividing point B, B < b, is impossible.  For, if I_c1, I_c2, . . . , I_cn is a set of intervals covering (a, B − δ), δ = δ_B/2, then  I_c1, I_c2, . . . , I_cn,   I_B   covers a ≤ x ≤ B + δ, so thatthere are accessible points to the right of B.  This is a contradiction, so that b must be accessible."
         −David Widder, Advanced Calculus; Dover Press, 1989; pg. 171

OBJECTIONS

1)  Consider the closed interval [a, b].  Let c = a.  Using Definition (1), To the point a of the closed interval a ≤ x ≤ b there corresponds a number δ_a and an interval I_a of length 2δ_a with a the center point,

     
We are in a dilemma.  There are two possibilities:
   A) δ_a is nonzero.   Specifically,
     (a − δ_a)< a  (I_a extends to the left of a) and
     the inequality (1) is satisfied.
   B) δ_a is zero, as in the example (2).

CASE A: Our cover may exceed the endpoints of [a, b].  Then we can also cover an open interval:
   Let δ_c be any function of c  and a such that lim_|c−a|→0(δ_c) > 0.

Clearly, we may do this with an offset.  What is not so clear, is that, if we must proceed by division... .well.  I think we need to have a talk about limits and power series.

CASE B:  This allows us to cover the interval [a, b] precisely.  However,  (1) cannot be satisfied:
      0 < 0 < 0  DNE.
We also lack a definition of the δ_c which can give us a finite cover and a zero radius at a.  I'm happy to make one up, and pick the δ_c however I please, but then I'll jus cover [a, b] with *cough*  I_c :  a ≤ x ≤ b.
We need to clarify how we intend, or are allowed, to derive δ_c .

If, in addition, our cover is not permitted to exceed the endpoints, we must approach both endpoints by division.   But then, as mentioned before, we cannot reach either endpoint by any finite process of division.  

2)  Every point of a bounded, open interval  a < x < b satisfies (1).   It is nevertheless not possible to precisely cover an open interval with a finite number of subintervals. To me, (1) is a working definition of open.  Every point in the interval is surrounded on both sides by other points of the interval.   Exactly two points in any closed interval fail the test: the endpoints.

3) Proof by contradiction requires us to establish that the two cases are mutually exclusive.  Widder argues that, given an interval I, if we can cover a ≤ x ≤ A  with a finite  number of strips of paper, and every point in the interval satisfies (1) then we have to be able to reach b with a finite number.  Is that true?  There are several issues here:
How did we get to A in the first place?
  i) We agreed to cover a ≤ x, ≤ A.  We have already thrown out open intervals by implicit assumption;  a is not in the interval (a, b).  All other criteria are now irrelevant and can be arbitrary.
 ii) We made up point A and assumed we can reach it.  That's silly.  This is the question we were supposed to answer:  Can this be assumed?  If it can be assumed for A it can be assumed for B.  Because we are begging the question. At this point we ought to just make up that we can reach b and be done with it.

It is possible to use this argument, but not the way it is presented, so I'd like to examine it:
 iii) If we can reach A, we can reach b.  We have restated the question, as a conclusion in the positive.  Can we fix it?
   Consider the closed interval [a, A].  Suppose a point β,   a < β < A,   is accessible from a. ...  We will then prove that A is accessible.  But we used A to prove that b was accessible!  If we insist on working this way, we will again require an infinite procedure to rediscover continuity: we will divide until our distance from a is zero, and then congratulate ourselves.  That's stupid and intellectually lazy.  The limit exists, we know it, and Widder takes it for granted without checking, but runs the argument as if the procedure is finite.

IF we proceed in this way, then the procedure is infinite.  We may, at any time, just lay off a finite number of intervals. There is no proof, any more than there was a proof that the points a, b and A, and the magic bridge  making A accessible from a exist.  They all arose by assumption.  A much simpler assumption is addition.  We get to the heart of the matter right away:
Choose an arbitrary point A in the closed interval [a, b], and a corresponding I_A.  Now OFFSET  I_A until its leftmost edge coincides with the leftmost point of the interval.
The distinction between an open and closed interval is immediately obtained.

Now.  To establish that reaching a from A, and failing to reach a from A are not mutually exclusive, let me add
iv)"Prove that some definite point A cannot be reached from a"
By assumption, each point of the interval a < x < A is the center of an interval (1).  Now, to ensure that we do not make any leaps, let us divide the distance A - a in such a way that we are certain the next accessible point lies in the range of the previous interval.
    Case 1:  Start at a:
    The interval  (a−δ_a ) < x < a either does not exist, or we have  a − δ_a ≤ x ≤ a, and δ_a = 0. In both cases, we cannot leave a.  There is no next interval.  We will obtain the same result starting from A, and measuring from there.  So,
    Case 2: Start at A, but measure from a:
  Choose the point c = 2(A− a)/3 + a, and let δ_c = (c − a)/2.    The range (c−δ_c) < x < (c+δ_c) is
     
Whew! Just made it.  Now, according to Widder, definition (1) renders it impossible, having reached c from A,  for there to exist a dividing point which prevents us from reaching a.
What could possibly go wrong?  Let let us continue.  The next interval has center at (2a +A)/3.
We have δ_c = (A−a)/6, and our new left endpoint (c−δ_c) is  (5a+A)/6.  Iterate again.  Now...
Let us say we reach a point arbitrarily close to a,  c = a+1/p,   p arbitrarily large.  We find that δ_c = 1/(2p), and the interval  (c−δ) < x <  (c+δ) is
     
 The dividing point which is supposed not to exist, is inescapably a.     We will never reach a.   We may now use a as the endpoint of an interval α < x < a, and find accessible points on the other side of a. Once again,  the open interval satisfies (1).   Given any point in the semi-open interval (a, b], we can name a value of c which contains the point, and a finite number of intervals which reach all the way to b.

  Of course, we are doing this to ourselves.  We are repeating Zeno's mistake.  It is essential to understand that the purple text is not a proof of anything.   It is the description of a choice: among the ways we may choose to measure, this one in particular has the stated properties.  When the argument is structured in a way that it admits all agreed methods of measure, and obtains conclusions which do not depend on the manner in which we approach it, then we can speak meaningfully of proof. Otherwise, we fail to distinguish the relationship among assumptions, facts, and argument.  Widder's error is informal: The assumption that there is a paradox which requires resolution.  There are as many ways to make the difficulty go away as we wish, all we have to do is pick one.  There is no logical proof.  For we may always proceed in this way, never-arriving.
.... and we may also always choose a larger interval.  For example, if we try I = [1, 3/2], and fail to arrive at a = 1 by division, we can instead use the interval [0, 2] and we will have no difficulty at all covering [1, 3/2] with a finite number of subdivisions.
... And we may also always skip the whole dillema and just divide the damn interval up however we want, slide subintervals left and right, and when we don't have defined endpoints infinity takes care of itself because we cannot name a leftmost and rightmost point.  DONE.

None of these are mutually contradictory.

4) He is everywhere taking continuity for granted.  GOOD.   But we all know where this is heading, right? He is going to play a little game where I am to pretend he is not taking continuity for granted, because he has never said it, and then continuity will spring forth from intervals by logical proof.

But an interval can't be defined without taking continuity for granted.  I mean,  literally, it is logically impossible.  There's no free lunch.  If you divide your way infinitely to zero from 1, it takes an infinite number of steps to get there, whether you are allowed to include the limit point or not.  If you let the cover exceed the whole interval, both open and closed intervals are covered; if you require the cover to reach the endpoint, the only way to get from one end to the other in a finite number of steps is by offset (addition and subtraction), in which case you need only ask "do I have endpoints to measure from?" in order to measure out a finite subdivision irrespective of the logic of this proof, or the properties of division.  The measure between need not even be continuous!  We imagine it so, when we have covered it, and the pretend picture in the mind is the only meaningful part of this proof.  If that's how I decide to accept continuity and identify all the singular points when I don't have it, fine.  I promise to have the courtesy not to tell you it's a proof.

(1) can't be satisfied by the endpoints of a closed interval.

←  Part I                                           Part III →

The Heine-Borel Theorem

Can't seem to make sense of the definition in my Calc book.  Time to lay it out.

Consider a finite number of points c_k on an interval I:
     
If I is an open interval (a,b), then  a < c₁,   and c_n < b.
If I is a closed interval [a,b], we may choose   c₁ = a    and/or    c₂ = b, but we do not have to.

Widder says:

6.1 THE HEINE-BOREL THEOREM
Let there correspond to each point c of the closed interval a ≤ x ≤ b a  number δ_c and an interval I_c of length 2δ_c with c the center point,

The Heine-Borel theorem states that a finite number of the intervals (1) can be chosen which will cover the whole interval a ≤ x ≤ b.  That is, every point of a ≤ x ≤ b will be in at least one of the above mentioned finite number of intervals (1).  In order to emphasize the need for proving this result, let us give an example to show it false if the interval (a, b) were open instead of closed.
    Let a = 0, b = 1, and define I_c, for 0 < c < 1, as

That is, δ_c = c/2.  No finite set of intervals (2) will cover the interval 0 < x < 1.  For, consider such a set, I_c1, I_c2, . . . , I_cn, where 0 < c₁ < c₂ . . . < c_n.  Of these, the interval I_c1  reaches farthest to the left.  Hence, no point to the left of c₁/2 is covered by the set.
                    − David Widder, Advanced Calculus; Dover Press,1989; pg 170

OBJECTIONS

1) But with δ_c= c/2, the closed interval [0, 1] cannot be covered by a finite number of intervals, either.  For consider a set   I_c1, I_c2, . . . , I_cn, where 0 = c₁ < c₂ . . . < c_n=1.  We cannot have c₁ = 0:
     
Which leaves us with:
    
And we cannot cover the range    0  ≤ x  <  c₁/2.

Allowing equalities at endpoints  does not ease the bind.  Let c₁ = 0.  Then we have
     
Now consider I_c2.  Its leftmost point is c₂/2.  We still cannot cover  0 < x <  c₂/2.

2) "Covered" remains undefined.

3) According to (1), each of the I_c has a nonzero length, 2δ_c.  If we allow the endpoints to be among the c_i, the "cover" can extend beyond the edges of the interval.  But if the cover may exceed the interval, we can cover an open interval as well: every point of (a, b) is covered by the same set that covers [a, b].
We may do this at any time by choosing a fixed width for δ_c.  Perhaps that is a trivial approach.  Perhaps Widder should specify what he expects of δ_c.

Note that, although we can approach 0 from above in such a way that we never arrive, division is bounded from below, and the limit exists:
       lim _{{n→ ∞}} 1/n  = 0.
 Enumeration is not bounded from above:
       lim_{{n → ∞}} n  = ∞.
{Oops, did I give away the mystery?}

If we do not allow ourselves to hold the limit point 0, and then ask how large we have to make n  to obtain zero, then of course there is no answer.

4) We can always use circular logic to prove our assumptions back to ourselves.
For suppose we cannot.
But this is contrary to our assumption, and so we can always use circular logic to prove our assumptions back to ourselves.

Widder measures differently when he has a closed, versus open, interval.  He uses offset (finite number of subintervals) when he has the limit point, and proceeds by infinite division when the limit cannot be used.
If we break his procedure up into distinct steps, he neglects to mention the steps in red for the closed interval:
    - subdivide any finite distance c − a
    - No finite number of divisions (intervals/inequalities) can obtain a.
    - By hypothesis, the limit exists (c − a is a finite quantity), so
    - Obtain the limit, a, by  infinite subdivision.
    - Is the limit a part of the interval?  
                           (yes)
    - agree to use it, instead of the limit sequence.
    - go back to ordinary measure and observe that we may lay off a finite number of subintervals between c and a.

The question in bold is where the two cases diverge.   In the middle of the argument, he asks if he can skip the proof altogether.  The answer is yes for the closed interval, no for the open.  Otherwise, his procedure is equally incapable of obtaining a closed interval in a finite number of steps.  The proof and definitions are irrelevant to the conclusion he obtains (see 5).

The problem remains: how ought they relate?

5) Suppose we choose our endpoint a arbitrarily small, but nonzero.  Leave δ_c,=c/2 as in the book example (2).  We get a different result.  The whole, open  interval a < x < b can be convered by a finite number of I_ck.  And likewise, a closed interval.
If the interval is open, let    a < c₁ < 3a/2.
  If the interval is closed,  we may use this same range for c₁.  In addition, we can let c₁=a, and the interval c₁/2 < x < 3c₁/2  is nonzero.

The procedure offered in (2) fails for any interval containing zero, where the width of the intervals I_ck  is some proper fraction of c.  Perhaps this is the result he wants.  He is about to use the Heine-Borel theorem to "prove" that a function which never escapes to infinity, over a finite range of x, has a *cough* upper limit.  
 But if, with a finite number of intervals of width 2δ_c,  where 2δ_c is a proper fraction of c, we can cover the entire range 0 ≤ x ≤ 1, then a finite number of intervals centered at c'_k = 1/c_k  and of width  δ'_c = 1/δ_c will cover the range 1 > x > ∞.  

6)  No interval containing 0 is compatible with both the definition (1) and the assumption of δ_c = c/2.  So there can be no subinterval centered at 0.  But in the following Theorem (see part 2), he requires that every point of the closed interval a ≤ x ≤ b  be the center of such a subinterval.  

7) We can obtain contrary results by choosing different functions for δ_c.  We have a situation where
Given ε there is a corresponding δ such that there are a finite number of ...
Given ε there is a corresponding δ such that there are not a finite number of...

Which is not a contradiction; δ_c is not unique or prescribed: we can measure in whatever fashion we wish. We may divide up a length by crossing half the distance between ourselves and the other side, and never arrive at the other side, as Xeno does, AND we can cross the whole distance in a single step.

Onward:                                                                                                   Part II: →

Continuity: Regions

IN THE BLUE CORNER, Advanced Calculus:

"

Multiple Integrals  

§ 1. Introduction 

In this chapter, we shall discuss double and triple integrals.  We shall follow as closely as possible the analogy with the theory of simple integrals developed in the previous chapter. 

1.1 REGIONS
     We have already discussed in Chapter 1 regions of the plane.  Let us collect here the notations which will be needed in the present chapter.
     A domain D is an open connected set of points.  That is, every point of D is the center of some δ-neighborhood, all of whose points are points of D; and any two points of D can be joined by a broken line having a finite number of segments, all of whose points are points of D.  A domain is bounded if all its points lie inside some square.
     A region R is a closed point set consisting of a bounded domain plus its boundary points.  We shall assume further that the boundary of R consists of a finite number of closed curves that do not cross themselves nor each other.  Note that the regions here defined and designated by the letter R are special cases of the more general ones of §3.1, Chapter 1.  In practical problems, R will usually be given in terms of its boundary curves.  For example, R might be the set of points between two concentric circumferences plus the points on the circumferences.  More frequently, we shall meet regions that can be most simply described by use of functions. Accordingly, we shall have a special notation for these.
     Let φ(x) and ψ(x) ∈ C in a ≤ x₁ ≤ b and φ(x) <ψ(x) in a < x < b.  Then the region R_x or R[a, b, φ(x), ψ(x)], is the region bounded by the curves   

x = a,     x = b,     y = φ(x),     y = ψ(x) 

If (x₁, y₁) is a point of R_x, then a ≤ x₁ ≤ b and φ(x₁)≤ y₁ ≤ ψ(x₁).  A line x = x₁, a  < x₁ < b cuts the boundary of R_x in just two points.  For example, the region R[-1, 1, − √(1 − x²), √(1 − x²)] is the circle x² + y² ≤ 1.  We could define in an obvious way a region R_y.  The region R described above as lying between two concentric circles is neither an R_x nor an R_y. It could be divided into four regions R_x, for example, by two vertical lines tangent to the inner circle.  These vertical lines would be counted twice, as the boundary of adjoining regions.
     A region R is simply connected if its boundary consists of a single closed curve.  The concept of the area of a region R will be assumed known.  Of course, the area of R_x is known from elementary calculus, and the area of R could be defined by use of a limiting process.  The diameter of a region R is the length of the longest line segment that joins two points of R.  In the case of a circle this coincides with the elementary notion of the diameter.  Observe that, if a region R varies so that its diameter approaches zero, then its area also approaches zero.  The converse is not true. 

1.2 DEFINITIONS
     We begin  by dividing a given region of R of area A into subregions.
As in the case of simple integrals, we introduce certain simplifying notations.
"
-Daivd Widder, Advanced Calculus

IN THE RED CORNER,  Euclid of Alexandria.

"The extremities of a surface are lines."
-Euclid, The Elements, Definition 6

_______________________________________ 

Notes

There are important differences between the two quotes.  We need the symbols necessary to operate algebraically, and to perform the Calculus.  To define arbitrary surfaces.  Our pencil is the function.  Still.  The surfaces we will be operating on in Chapter 6 of Widder's Advanced Calculus are precisely those which are defined in Euclid.  The definition is rigorous.  And so,

I have some questions for you, Modern Abstract Algebra.  And I don't mean Gauss.

1. What is the difference between a Dedekind section and Euclid's Definitions?
2. What is the relationship between infinite division and continuous measure?
   (Hint: No matter what anyone says, I can prove false by counterexample any mathematical argument which concludes by proof that I, Ryan, can measure.  All I need to do is disagree.  Likewise, a proof that I cannot measure.  All I need to do is say, "one apple."
   And likewise, I can prove any assertion that integral, unbroken measure does not exist, by waiting for a new instant in time, unbroken from the previous.  Or taking a step, some distance in some direction.  Left foot, right foot.  I defeat you, fake divisor-paradox!  Time for a nap.)
3. What is the logical status of a postulate?
4. I say, mathematics is not a logical circle.  Why?

I think, Mr. Modern, you do not answer these questions accurately.  Prove me wrong.  I'm asking for it.  See?

{Thus goad I a concept.
 Poke. 
 Thus}

Double integration equivalence

Time to clarify my assumptions about integration.

The double integral,

 is shorthand for:
             

We begin at the innermost integral.   Integrate with respect to x, holding  y constant.  Then integrate the result with respect to y (holding x constant).  Each step is an ordinary, single-variable integral.  We may also change the order of integration:
 

Just remember to keep x-bounds with x, and  y-bounds with y.

These are convenient results, and after hammering through partial differentiation, integrating in any order makes intuitive sense.  This is the next chapter of Advanced Calculus, but I think the result is self-evident, and I want to clarify that assumption so I can find any mistakes.

  I have heard equality (2)  referred to as Fubini's Theorem, but the statement before me is, the two integrals are equivalent.  The only Theorem I have any business with must be the Fundamental Theorem of the Calculus.  If I break up (2) into infinite sums, the sum on the right side of the equal sign must be the same as that on the left.  Right?

Variable bounds, paramaterization, partial summation with discrete and continuous quantities are a few pages away.  I  need a working, base example: a continuous function with constant bounds.  Let me build that now with the simplest case: a dobule integral of the form (1), with constant bounds.

A FUNCTION OF A SINGLE VARIABLE

Let F(x) be a function whose derivative, F'(x) = f(x)  is continuous over the closed interval [a, b], i.e.
              
Then the Fundamental Theorem of Calculus states


PROOF:  This approach is abstract, but concise.  By continuity, we can write the integral as the limit of a sum


Where δ is the largest subinterval and ξ_k lies in the subinterval [x_k - x_k-1].
Now, we choose for our  ξ_k the points in each subinterval where the line tangent to F(x)  has the same slope as the segment joining the endpoints, F(x_k ) and  F(x_k-1):

             (Rise/Run) (Run)             =   (Rise)

Then the right hand side of (4) becomes,


Rearranging the terms,  we have

    _____________________________________________________


And the theorem is proved.  Let me try to extend this method myself.

A FUNCTION OF TWO INDEPENDENT VARIABLES

Beginning with equation (1):
     
First, consider the inner integral:
     
What relationship does f(x,y)dx have to F(x)?  It is the partial derivative F₂₁(x):
     

The result of the first integration will then be  F₂(x).  This is my principal assumption. Specifically,
     
I'm not sure how I feel about that.  The assumption is sufficient to make the sums work out.  And to use the Fundamental Theorem, we must already know the relationship between f and F.  I don't see a way around equality (8).   But I already know that F₂₁(x, y,) and F₁₂(x, y) are identical for 'nice' functions, and the FTC is just going to give that information back to me.

That's why  I need to work out the concrete terms, determine if it holds, and if so, how.  Otherwise I'm thinking in a circle.  I will use the notation of partial derivatives, but I did not begin with this assumption.  I tried several ways of indexing the sums, and in the end found that "indexing" was the same thing as labeling partial derivatives.  The FTC makes this process clear.

I.  Integrate with respect to x
     

Choose the ξ_k so that
     
This is a directional derivative.  There is no slope in the y direction because I'm not varying y. Hence the equality.  This is the crux.  Not-varying one of two independent variables is exactly the same thing as taking a partial derivative. The x intervals sum and cancel to a discrete value, independently of the y intervals.  Multiplication and addition are commutative and associative, so the sums and products of these values can be arranged in any order.  Taking the integrals separately is also exactly the same thing as saying the integration can be done in any order. x and y never mix in the infinite sums.  That's not a theorem; that's what we're doing.  This is what I mean about Fubini's theorem, whatever the hell it is.  Integrating the independent variables separately is how we insure there is only one answer.  Onward!

Now we have


Rearranging the terms,

   __________________________________________________________________



II.  Integrate with respect to  y
I will integrate the general form:
     
For the purposes of integration, this is the same as
     
and can be applied to (7)'.   The process is now the same as above.
     
Choose the ζ_k  so that
     

That's still a derivative on the left! (5)'' can alternately read,
                 
Thus we have


And the last line can again be rearranged so that all but two terms cancel:

   ______________________________________________________________



III. Equivalence
Now compute the result:


... with the happy result that giving one order of integration entails the equivalence of the other.

HOW'S MY DRIVING?

   I have the feeling double integration is applicable in a variety of circumstances, leaving this approach insufficient for many cases.  Or maybe I made mistakes, and it's all wrong.  Somebody is letting a 12-year old drive a semi, and both of those guys are me.  Tell me how my driving is.  Leave me a comment.  Correct a mistake.  Prattle on about Space-Wogs.

Fuckin' space wogs.  Always framping the scrumpets.

{Thanks again to CodeCogs for free LaTeX equation editing and hosting the resulting, tiny bitches images.}