The Heine-Borel Theorem, 2

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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.

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