{CLEANUP TO FOLLOW}
I swear it's got to be in here somewhere. I think it's safe to move on now.
I say:
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.
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.
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.
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.
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 ci.
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 δc:
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 δc > 0 which is
the closest point to c, so that only c lives
inside the circle.
Then
you name ε = δc/2, and 0 < ε < δc,
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−δc,
c+δc], 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.
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 Ik, for both closed and open endpoint. The same cannot be said for an endpoint open at infinity.
- . . .Where
we want . . . *δc to
be a function of c. *To
use the same definition when the interval
involves zero.
*To distinguish limits with intervals.
What is a correct statement of the Heine-Borel Theorem?
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