Problem with my intuition (measure theory)
none
I have a problem with my intuition. Maybe someone out there can help me
understand what's wrong with it. (Maybe I make some erroneous assumption
I'm not aware of?)
The following contradictory "proof" illustrates the problem:
Claim: The Lebesgue measure of the unit square is zero.
"Proof": Let {qn} be a countable enumeration of all points in the square
with rational coordinates. Cover each point qn with an open square
centered in qn and let it have area (e/(e+1))^k, with e>0. Then the sum
of the area of the squares is the sum of the geometric progession
(e/(e+1))^k where k=1,2,... This sum is e.
Suppose this set of squares cover the unit square. Then the Lebesgue
measure is the infimum of the area of the cover, which is clearly zero.
Hence it cannot be a cover.
But how can it not be? All the squares has non-zero area, the set {qn}
is dense in the unit square and lies closer to each other than any real
number.
Can anyone name a point which is not covered?
Any help is much appreciated! / Regards, Daniel
Don Coppersmith
Make a similar argument in one dimension, where it's
easier to compute.
Start with the closed interval [0,2].
Centered around each rational p/q,
put an open interval of length 1/(10q^2).
Now we can argue that sqrt(2) is not covered by
any of these open intervals.
If it were, then we woulod have
| p/q - sqrt(2) | < 1/(10q^2).
But we know (p/q + sqrt(2)) < 4.
Multiplying,
| p^2/q^2 - 2 | < 4/(10q^2).
But p^2 is unequal to 2q^2,
so | p^2/q^2 - 2 | is nonzero,
and it's a multiple of 1/q^2, so it is at least 1/q^2.
Contradiction.
So sqrt(2) is not covered by any of these intervals.
The same thing goes on in your 2-dimensional example;
it's just harder to write an explicit example.
Don Coppersmith
Arturo Magidin
>Hello!
>
>I have a problem with my intuition. Maybe someone out there can help me
>understand what's wrong with it. (Maybe I make some erroneous assumption
>I'm not aware of?)
>
>The following contradictory "proof" illustrates the problem:
>
>Claim: The Lebesgue measure of the unit square is zero.
>
>"Proof": Let {qn} be a countable enumeration of all points in the square
>with rational coordinates. Cover each point qn with an open square
>centered in qn and let it have area (e/(e+1))^k, with e>0. Then the sum
>of the area of the squares is the sum of the geometric progession
>(e/(e+1))^k where k=1,2,... This sum is e.
Huh? The geometric series has sum
1/(1 - (e/e+1)) = 1/(1/(e+1)) = e+1
not e.
>Suppose this set of squares cover the unit square. Then the Lebesgue
>measure is the infimum of the area of the cover, which is clearly zero.
>Hence it cannot be a cover.
No. The infimum is 1.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
none
> In article <4338483c$1...@griseus.its.uu.se>, none <""john\"@(none)"> wrote:
>
>>Hello!
>>
>>I have a problem with my intuition. Maybe someone out there can help me
>>understand what's wrong with it. (Maybe I make some erroneous assumption
>>I'm not aware of?)
>>
>>The following contradictory "proof" illustrates the problem:
>>
>>Claim: The Lebesgue measure of the unit square is zero.
>>
>>"Proof": Let {qn} be a countable enumeration of all points in the square
>>with rational coordinates. Cover each point qn with an open square
>>centered in qn and let it have area (e/(e+1))^k, with e>0. Then the sum
>>of the area of the squares is the sum of the geometric progession
>>(e/(e+1))^k where k=1,2,... This sum is e.
>
>
> Huh? The geometric series has sum
>
> 1/(1 - (e/e+1)) = 1/(1/(e+1)) = e+1
>
> not e.
I believe the you are summing from k=0, instead of k=1 as specified. I'm
sorry if I was not clear. In any case, you get
sum = 1/(1 - (e/e+1)) - 1 = 1/(1/(e+1)) - 1 = e+1-1 = e
/ Daniel
C6...@shaw.ca
> Hello!
>
> I have a problem with my intuition. Maybe someone out there can help me
> understand what's wrong with it. (Maybe I make some erroneous assumption
> I'm not aware of?)
>
> The following contradictory "proof" illustrates the problem:
>
> Claim: The Lebesgue measure of the unit square is zero.
>
> "Proof": Let {qn} be a countable enumeration of all points in the square
> with rational coordinates. Cover each point qn with an open square
> centered in qn and let it have area (e/(e+1))^k, with e>0. Then the sum
> of the area of the squares is the sum of the geometric progession
> (e/(e+1))^k where k=1,2,... This sum is e.
>
> Suppose this set of squares cover the unit square. Then the Lebesgue
> measure is the infimum of the area of the cover, which is clearly zero.
> Hence it cannot be a cover.
>
> But how can it not be? All the squares has non-zero area, the set {qn}
> is dense in the unit square
I think there is a difference between saying that every real is close
to some qn (which is "densness") vs. saying that every qn is close to
some real. For any real point and any neighbourhood of diamater
epsilon, there exist some qn lying in that neighbourhood, so we are
essentially fixing the real and varying the qn. Your argument is the
opposite: it fixes the qn and varies the real, and I guess that is the
source of the trouble.
Ray Vickson
Arturo Magidin
Fair enough. As was pointed out, your error is that this sequence of
neighborhoods does NOT cover the unit square. Just because for every
x, for every n, there is are rationals in a neighborhood centered at x
with total area (e/e+1)^n, it does not follow that your neighborhoods
cover all the square: all the rationals in that neighborhood may have
indices too large to include x in their corresponding neighborhod.
For example, imagine taking small neighborhoods around the points 1/n
in the real line. You can pick a neighborhood of size 1/2^n around
1/n, and none of these neighboorhoods will include 0, even though for
every e>0 there are an infinite number of points of the form 1/n in
the neighborhood (-e, e) of 0.
David C. Ullrich
Of course nobody can name such a point, since you haven't
specified exactly what the covering is.
But one can easily show that there exists an example of
what you did where one _can_ name such a point. _Fix_
a point p in the square with irrational coordinates.
Now think about your construction: Each time you choose
a square covering qn, make certain that the square is
small enough that it does not cover p. Then p is not
covered by any square...
>Any help is much appreciated! / Regards, Daniel
************************
David C. Ullrich
Keith Ramsay
none wrote:
|But how can it not be? All the squares has non-zero area, the set {qn}
|is dense in the unit square and lies closer to each other than any
real
|number.
It's funny how intuition can go wrong here, isn't it? I suspect
that it comes from thinking of this as being more like the case
where you surround each qn with a neighborhood *of the same size*
than it is. If the epsilon is independent of n, then given a
point P, we can just pick a qn that's within epsilon of P, and its
neighborhood includes P.
Here the nonuniformity of the sizes of the neighborhoods is key.
For each epsilon>0, there is some qn that lies within epsilon of
P; also for each qn, there is some epsilon>0 so that the points
within epsilon of qn are covered. But the rate at which the needed
n increases as epsilon goes to 0, and the rate at which the epsilon
decreases with increasing n, don't "match up".
Suppose we took just a 1-dimensional example. For each rational
p/q, take the neighborhood of radius 1/10q^3 around it. This may
sound silly, but picture it like this: imagine that the point
p/q (in lowest terms) is at a depth of q below some level, so that
as the rationals get more complicated, they go deeper and deeper
below the ground. Further imagine this interval (p/q-1/10q^3,
p/q+1/10q^3) as being a hill with its top at the rational p/q,
and sloping down to -infinity at the endpoints of the interval.
So basically I'm suggesting you picture the graph of a function
which is the maximum over p/q of some function with a value -q
at p/q and going to -infinity approaching p/q+-1/10q^3.
Now imagine being at a point like (sqrt(5)-1)/2, the golden mean.
Start at height 0 and start falling. The highest hills are at
the integers; the hill at 1 is just a depth of 1, but it's narrow
enough that you miss it. It only extends to 9/10 but the golden
mean is between 6/10 and 7/10. Then there's a hill at 1/2 at a
depth of 2. But it's narrower and we miss it too.
As we continue to fall, we encounter rationals closer and closer to
(sqrt(5)-1)/2. But the hills they are on are getting narrower and
narrower too. The real question is, how quickly are they getting
narrow, compared to how quickly they are getting close. In this case
it turns out that they aren't getting close fast enough. We keep
falling forever, having fallen in a "crack" between the rational
numbers. In the case of this particular number, the closest rational
approximations satisfying a bound on the denominator are the ones
one gets from the Fibonacci sequence:
1/1, 1/2, 2/3, 3/5, 5/8, 8/13, ...
which are getting closer like one over the square of the denominator.
But I chose to make the size of the neighborhood around each one get
smaller in proportion to one over the cube of the denominator, which
is going to zero enough faster to keep up.
This kind of infinite race where something is tending to zero but
not fast enough to catch up with something else, can only occur
because we're dealing with infinite sequences. It's a pretty common
occurrence to find some form of it counterintuitive at least to
begin with. I think some professionals still find dense sets having
measure zero a little funny. But as you get familiar with this kind
of situation the sense of paradox tends to wear off.
Keith Ramsay
Ross A. Finlayson
And for irrationals 1/en and intervals 1/en^2?
Re-Vitali-ize measure theory. Well-order the reals.
Ross